NCERT Solutions Class 11 Chapter 12 Limits And Derivatives Miscellaneous Exercise (original) (raw)

Last Updated : 22 Apr, 2024

Question 1: Find the derivative of the following functions from first principle:

(i) -x

**Solution:

f(x) = -x

f(x+h) = -(x+h)

From the first principle,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{-(x+h)-(-x)}{h})\\ f'(x) = \lim_{h \to 0} (\frac{-x-h+x}{h})\\ f'(x) = \lim_{h \to 0} (\frac{-h}{h})\\ f'(x) = \lim_{h \to 0} -1

f'(x) = -1

(ii) (-x)-1

**Solution:

f(x) = (-x)-1 = \frac{-1}{x}

f(x+h) = (-(x+h))-1 = \frac{-1}{x+h}

From the first principle,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{\frac{-1}{x+h}-(\frac{-1}{x})}{h})\\ f'(x) = \lim_{h \to 0} (\frac{\frac{-1}{x+h}+\frac{1}{x}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{\frac{-x+(x+h)}{x(x+h)}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{h}{hx(x+h)})\\ f'(x) = \lim_{h \to 0} (\frac{1}{x(x+h)})\\ f'(x) = (\frac{1}{x(x+0)})\\ f'(x) = (\frac{1}{x^2})

(iii) sin(x+1)

**Solution:

f(x) = sin(x+1)

f(x+h) = sin((x+h)+1)

From the first principle,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{sin((x+h)+1)-(sin(x+1))}{h})

Using the trigonometric identity,

**sin A - sin B = 2 cos (\frac{A+B}{2}) sin (\frac{A-B}{2})

f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{x+h+1+(x+1)}{2}) sin (\frac{x+h+1-(x+1)}{2}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{2x+h+2}{2}) sin (\frac{h}{2})}{h})\\ f'(x) = \lim_{h \to 0} (2 cos (\frac{2x+h+2}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h}))

Multiply and divide by 2, we have

f'(x) = 2 cos (\frac{2x+0+2}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) \times \frac{2}{2}\\ f'(x) = 2 cos (x+1) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{\frac{h}{2}})) \times \frac{1}{2}

f'(x) = cos (x+1) (1)

f'(x) = cos (x+1)

(iv) cos(x-\frac{\pi}{8})

**Solution:

Here, f(x) = cos(x-\frac{\pi}{8})\\ f(x+h) = cos((x+h)-\frac{\pi}{8})

From the first principle,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{cos((x+h)-\frac{\pi}{8})-cos(x-\frac{\pi}{8})}{h}

Using the trigonometric identity,

**cos a - cos b = -2 sin (\frac{a+b}{2}) sin (\frac{a-b}{2})

f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{(x+h)-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}) sin (\frac{(x+h)-\frac{\pi}{8}-(x-\frac{\pi}{8})}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{(2x+h)-\frac{2\pi}{8}}{2}) sin (\frac{(x+h)-\frac{\pi}{8}-x+\frac{\pi}{8})}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{2x+h-\frac{\pi}{4}}{2}) sin (\frac{h}{2})}{h}

Multiplying and dividing by 2,

f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{2x+h-\frac{\pi}{4}}{2}) sin (\frac{h}{2})}{h} \times \frac{2}{2}\\ f'(x) = \lim_{h \to 0} (-2 sin (\frac{2x+h-\frac{\pi}{4}}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2}\\ f'(x) = (-sin (\frac{2x+0-\frac{\pi}{4}}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}}\\ f'(x) = (-sin (x-\frac{\pi}{8})) (1)\\ f'(x) = -sin (x-\frac{\pi}{8})

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

Question 2: (x+a)

**Solution:

f(x) = x+a

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(x+a)\\ f'(x) = \frac{d}{dx}(x)+\frac{d}{dx}(a)

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (1.x^{1-1})+0\\ f'(x) = (1.x^0)+0\\ f'(x) = 1

Question 3: (px+q) (\frac{r}{s}+s)

**Solution:

f(x) = (px+q) (\frac{r}{s}+s)

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}((px+q) (\frac{r}{s}+s))

Using the product rule, we have

****(uv)' = uv'+u'v**

f'(x) = (px+q) \frac{d}{dx}(\frac{r}{s}+s) + (\frac{r}{s}+s) \frac{d}{dx}(px+q)

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (px+q) (0) + (\frac{r}{s}+s) (\frac{d}{dx}(px)+\frac{d}{dx}(q))\\ f'(x) = 0 + (\frac{r}{s}+s) (p+0)\\ f'(x) = p(\frac{r}{s}+s)

Question 4: (ax+b) (cx+d)2

**Solution:

f(x) = (ax+b) (cx+d)2

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}((ax+b) (cx+d)^2)

Using the product rule, we have

****(uv)' = uv'+u'v**

f'(x) = (ax+b) \frac{d}{dx}(cx+d)^2 + (cx+d)^2 \frac{d}{dx}(ax+b)

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (ax+b) \frac{d}{dx}(c^2x^2+2(cx)(d)+d^2) + (cx+d)^2 (\frac{d}{dx}(ax)+\frac{d}{dx}(b))\\ f'(x) = (ax+b) (c^2(2x)+2cd+0) + (cx+d)^2 (a+0)\\ f'(x) = (ax+b) (2xc^2+2cd) + a(cx+d)^2

Question 5: \frac{ax+b}{cx+d}

**Solution:

f(x) = \frac{ax+b}{cx+d}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{ax+b}{cx+d})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(cx+d) \frac{d}{dx}(ax+b) - (ax+b)\frac{d}{dx}(cx+d)}{(cx+d)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(cx+d) (a) - (ax+b)(c)}{(cx+d)^2}\\ f'(x) = \frac{acx+ad - acx-bc}{(cx+d)^2}\\ f'(x) = \frac{ad-bc}{(cx+d)^2}

Question 6: \frac{1+\frac{1}{x}}{1-\frac{1}{x}}

**Solution:

f(x) = \frac{1+\frac{1}{x}}{1-\frac{1}{x}}\\ f(x) = \frac{\frac{x+1}{x}}{\frac{x-1}{x}}\\ f(x) = \frac{x+1}{x-1}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x+1}{x-1})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(x-1) \frac{d}{dx}(x+1) - (x+1)\frac{d}{dx}(x-1)}{(x-1)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(x-1) (1) - (x+1)(1)}{(x-1)^2}\\ f'(x) = \frac{(x-1 - x-1)}{(x-1)^2}\\ f'(x) = \frac{-2}{(x-1)^2}\\

Question 7: \frac{1}{ax^2+bx+c}

**Solution:

f(x) = \frac{1}{ax^2+bx+c}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{1}{ax^2+bx+c})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(ax^2+bx+c) \frac{d}{dx}(1) - (1)\frac{d}{dx}(ax^2+bx+c)}{(ax^2+bx+c)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(ax^2+bx+c) (0) - (1)(a(2x)+b+0)}{(ax^2+bx+c)^2}\\ f'(x) = \frac{0 - (2ax+b+0)}{(ax^2+bx+c)^2}\\ f'(x) = \frac{- (2ax+b+0)}{(ax^2+bx+c)^2}

Question 8: \frac{ax+b}{px^2+qx+r}

**Solution:

f(x) = \frac{ax+b}{px^2+qx+r}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{ax+b}{px^2+qx+r})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(px^2+qx+r) \frac{d}{dx}(ax+b) - (ax+b)\frac{d}{dx}(px^2+qx+r)}{(px^2+qx+r)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(px^2+qx+r) (a+0) - (ax+b)\frac{d}{dx}(p(2x)+q+0)}{(px^2+qx+r)^2}\\ f'(x) = \frac{a(px^2+qx+r) - (ax+b)(2px+q)}{(px^2+qx+r)^2}\\ f'(x) = \frac{apx^2+qax+ra - (2apx^2+qax+2bpx+bq)}{(px^2+qx+r)^2}\\ f'(x) = \frac{apx^2+qax+ra - 2apx^2-qax-2bpx-bq)}{(px^2+qx+r)^2}\\ f'(x) = \frac{ra - apx^2-2bpx-bq)}{(px^2+qx+r)^2}

Question 9: \frac{px^2+qx+r}{ax+b}

**Solution:

f(x) = \frac{px^2+qx+r}{ax+b}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{px^2+qx+r}{ax+b})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(ax+b) \frac{d}{dx}(px^2+qx+r) - (px^2+qx+r)\frac{d}{dx}(ax+b)}{(ax+b)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(ax+b) (p(2x)+q+0) - (px^2+qx+r)(a+0)}{(ax+b)^2}\\ f'(x) = \frac{(ax+b) (2px+q) - (px^2+qx+r)(a)}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - (apx^2+qax+ra}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - apx^2-qax-ra}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - apx^2-qax-ra}{(ax+b)^2}\\ f'(x) = \frac{apx^2 - ra +2bpx+bq)}{(ax+b)^2}

Question 10: \frac{a}{x^4}-\frac{b}{x^2}+cos x

**Solution:

f(x) = \frac{a}{x^4}-\frac{b}{x^2}+cos x

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{a}{x^4}-\frac{b}{x^2}+cos x)\\ f'(x) = \frac{d}{dx}(\frac{a}{x^4})-\frac{d}{dx}(\frac{b}{x^2})+\frac{d}{dx}(cos x)

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = a \frac{d}{dx}(x^{-4}) - b\frac{d}{dx}(x^{-2})+(- sin x)\\ f'(x) = a (-4x^{-4-1}) - b((-2)x^{-2-1})+(- sin x)\\ f'(x) = a (-4x^{-5}) - b((-2)x^{-3})+(- sin x)\\ f'(x) = -[4ax^{-5} - 2bx^{-3} + sin x]

Question 11: 4\sqrt{x} - 2

**Solution:

f(x) = 4\sqrt{x} - 2

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(4\sqrt{x} - 2)\\ f'(x) = \frac{d}{dx}(4\sqrt{x})-\frac{d}{dx}(2)

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = 4 \frac{d}{dx}(x^{\frac{1}{2}})-0\\ f'(x) = 4 (\frac{1}{2}x^{\frac{1}{2}-1})\\ f'(x) = 2 (x^{-\frac{1}{2}})\\ f'(x) = \frac{2}{\sqrt{x}}

Question 12: (ax+b)n

**Solution:

f(x) = (ax+b)n

f(x+h) = (a(x+h)+b)n

f(x+h) = (ax+ah+b)n

From the first principle,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-(ax+b)^n}{h}\\ f'(x) = \lim_{h \to 0} \frac{(ax+ah+b)^n-(ax+b)^n}{h}\\ f'(x) = \lim_{h \to 0} \frac{(ax+b)^n (1+\frac{ah}{ax+b})^n-(ax+b)^n}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} \frac{(1+\frac{ah}{ax+b})^n-1}{h}

Using the binomial expansion, we have

f'(x) = (ax+b)^n \lim_{h \to 0} \frac{[1+n(\frac{ah}{ax+b}+\frac{n(n-1)}{2!}(\frac{ah}{(ax+b)^2}+.....]-1}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} \frac{[n(\frac{ah}{ax+b}+\frac{n(n-1)a^2h^2}{2!(ax+b)^2}+.................]}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} [n(\frac{a}{ax+b}+\frac{n(n-1)a^2h}{2!(ax+b)^2}+.................]\\ f'(x) = (ax+b)^n [n(\frac{a}{ax+b}+\frac{n(n-1)a^2(0)}{2!(ax+b)^2}+0+0+0......]\\ f'(x) = (ax+b)^n (\frac{na}{ax+b})\\ f'(x) = na (ax+b)^{n-1}

Question 13: (ax+b)n (cx+d)m

**Solution:

f(x) = (ax+b)n (cx+d)m

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}((ax+b)^n (cx+d)^m)

Using the product rule, we have

(uv)' = uv'+u'v

f'(x) = (ax+b)^n \frac{d}{dx}(cx+d)^m + (cx+d)^m \frac{d}{dx}(ax+b)^n

Let's take, g(x) = (cx+d)m

g'(x) = \frac{d}{dx}(cx+d)^m

g(x+h) = (c(x+h)+d)m

g(x+h) = (cx+ch+d)m

From the first principle,

g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{(cx+ch+d)^m-(cx+d)^m}{h}\\ g'(x) = \lim_{h \to 0} \frac{(cx+d)^m (1+\frac{ch}{cx+d})^m-(cx+d)^m}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} \frac{(1+\frac{ch}{cx+d})^m-1}{h}

Using the binomial expansion, we have

g'(x) = (cx+d)^m \lim_{h \to 0} \frac{[1+m(\frac{ch}{cx+d}+\frac{m(m-1)}{2!}(\frac{ch}{(cx+d})^2+.....)]-1}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} \frac{[m(\frac{ch}{cx+d}+\frac{m(nm-1)c^2h^2}{2!(cx+d)^2}+.................)]}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} [m(\frac{c}{cx+d}+\frac{m(m-1)c^2h}{2!(cx+d)^2}+.................)]\\ g'(x) = (cx+d)^m [m(\frac{c}{cx+d}+\frac{m(m-1)c^2(0)}{2!(cx+d)^2}+0+0+0......)]\\ g'(x) = (cx+d)^m (\frac{mc}{cx+d})\\ g'(x) = mc (cx+d)^{m-1}

So, as

f'(x) = (ax+b)^n \frac{d}{dx}(cx+d)^m + (cx+d)^m \frac{d}{dx}(ax+b)^n\\ f'(x) = (ax+b)^n (mc (cx+d)^{m-1}) + (cx+d)^m (na (ax+b)^{n-1})

Question 14: sin (x + a)

**Solution:

f(x) = sin(x+a)

f(x+h) = sin((x+h)+a)

From the first principle,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{sin((x+h)+a)-(sin(x+a))}{h})

Using the trigonometric identity,

**sin A - sin B = 2 cos (\frac{A+B}{2}) sin (\frac{A-B}{2})

f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{x+h+a+(x+a)}{2}) sin (\frac{x+h+a-(x+a)}{2}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{2x+h+2a}{2}) sin (\frac{h}{2})}{h})\\ f'(x) = \lim_{h \to 0} (2 cos (\frac{2x+h+2a}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h}))

Multiply and divide by 2, we have

f'(x) = 2 cos (\frac{2x+0+2a}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) \times \frac{2}{2}\\ f'(x) = 2 cos (x+a) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{\frac{h}{2}})) \times \frac{1}{2}\\ f'(x) = cos (x+a) (1)\\ f'(x) = cos (x+a)

Question 15: cosec x cot x

**Solution:

f(x) = cosec x cot x

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(cosec x \hspace{0.1cm}cot x)

Using the product rule, we have

****(uv)' = uv'+u'v**

f'(x) = cot x \frac{d}{dx}(cosec x) + (cosec x) \frac{d}{dx}(cot x)

f'(x) = cot x (-cot x cosec x) + (cosec x) (-cosec2 x)

f'(x) = - cot2 x cosec x - cosec3 x

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

Question 16: \frac{cos x}{1+sin x}

**Solution:

f(x) = \frac{cos x}{1+sin x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{cos x}{1+sin x})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(1+sin x) \frac{d}{dx}(cos x) - (cos x)\frac{d}{dx}(1+sin x)}{(1+sin x)^2}\\ f'(x) = \frac{(1+sin x) (-sin x) - (cos x)(0+cos x)}{(1+sin x)^2}\\ f'(x) = \frac{(-sinx - sin^2 x)- (cos^2 x)}{(1+sin x)^2}\\ f'(x) = \frac{-sinx - (sin^2 x + cos^2 x)}{(1+sin x)^2}\\ f'(x) = \frac{-sinx - 1}{(1+sin x)^2}\\ f'(x) = \frac{-(sinx + 1)}{(1+sin x)^2}\\ f'(x) = \frac{-1}{(1+sin x)}

Question 17: \frac{sin x + cos x}{sin x - cos x}

**Solution:

f(x) = \frac{sin x + cos x}{sin x - cos x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{sin x + cos x}{sin x - cos x})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(sin x - cos x) \frac{d}{dx}(sin x + cos x) - (sin x + cos x)\frac{d}{dx}(sin x - cos x)}{(sin x - cos x)^2}\\ f'(x) = \frac{(sin x - cos x) (cos x + (- sin x)) - (sin x + cos x)(cos x - (-sin x))}{(sin x - cos x)^2}\\ f'(x) = \frac{(sin x - cos x) (cos x - sin x) - (sin x + cos x)(cos x + sin x)}{(sin x - cos x)^2}\\ f'(x) = \frac{- (sin x - cos x) (sin x - cos x) - (sin x + cos x)(cos x + sin x)}{(sin x - cos x)^2}\\ f'(x) = - \frac{(sin x - cos x)^2 + (sin x + cos x)^2}{(sin x - cos x)^2}\\ f'(x) = - \frac{(sin^2 x - 2 sin x cos x + cos^2 x) + (sin^2 x - 2 sin x cos x + cos^2 x)}{(sin x - cos x)^2}\\ f'(x) = - \frac{1+1}{(sin x - cos x)^2}\\ f'(x) = \frac{-2}{(sin x - cos x)^2}

Question 18: \frac{sec x - 1}{sec x + 1}

**Solution:

f(x) = \frac{sec x - 1}{sec x + 1}\\ f(x) = \frac{\frac{1}{cos x} - 1}{\frac{1}{cos x} + 1}\\ f(x) = \frac{\frac{1-cos x}{cos x}}{\frac{1+cos x}{cos x}}\\ f(x) = \frac{1-cos x}{1+cos x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{1-cos x}{1+cos x})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(1+cos x) \frac{d}{dx}(1-cos x) - (1-cos x)\frac{d}{dx}(1+cos x)}{(1+cos x)^2}\\ f'(x) = \frac{(1+cos x) (0-(- sin x)) - (1-cos x)(0+(- sin x))}{(1+cos x)^2}\\ f'(x) = \frac{(1+cos x) (sin x) - (1-cos x)(- sin x)}{(1+cos x)^2}\\ f'(x) = \frac{(sin x + sin x \hspace{0.1cm}cos x) + (sin x - sin x\hspace{0.1cm} cos x)}{(1+cos x)^2}\\ f'(x) = \frac{(sin x + sin x)}{(1+cos x)^2}\\ f'(x) = \frac{2 sin x}{(1+cos x)^2}

Question 19: sinn x

**Solution:

f(x) = sinn x

**When n = 1,

f(x) = sin x

f'(x) = \frac{d}{dx}(sin x) = cos x

**When n = 2,

f(x) = sin2 x = sin x sin x

f'(x) = \frac{d}{dx}(sin x\hspace{0.1cm} sin x)

Using the product rule, we have

****(uv)' = uv'+vu'**

f'(x) = (sin x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin x)

f'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

f(x) = sin3 x = sin2 x sin x

f'(x) = \frac{d}{dx}(sin^2\hspace{0.1cm} x sin x)

Using the product rule, we have

****(uv)' = uv'+vu'**

f'(x) = (sin^2 x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin^2 x)\\ f'(x) = (sin^2 x) (cos x) + (sin x) (2 sin x\hspace{0.1cm} cos x)\\ f'(x) = (sin^2 x \hspace{0.1cm}cos x) + (2 sin^2 x \hspace{0.1cm}cos x)\\ f'(x) = (sin^2 x\hspace{0.1cm} cos x)[1+2]\\ f'(x) = 3 sin^2 x \hspace{0.1cm}cos x

Pattern w.r.t n is seen here, as follows

\frac{d}{dx}(sin^n x) = n sin^{n-1}x\hspace{0.1cm} cos x

Let's check this statement.

For P(n) = n sinn-1x cos x

For P(1),

P(1) = 1 sin1-1x cos x = cos x. Which is true.

n=k

P(k) = k sin^{k-1}x \hspace{0.1cm}cos x

n = k+1

P(k+1) = \frac{d}{dx}(sin^{k+1} x) = \frac{d}{dx}(sin^k x \hspace{0.1cm}sin x)

Using the product rule, we have

****(uv)' = uv'+vu'**

= (sink x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin^k x)

= (sink x) (cos x) + (sin x) (k sink-1 x cos x)

= (sink x) (cos x)[k+1]

Hence proved for P(k+1).

So, \frac{d}{dx}(sin^n x) = n sin^{n-1}x \hspace{0.1cm}cos x is true.

Question 20: \frac{a+bsin x}{c+dcos x}

**Solution:

f(x) = \frac{a+bsin x}{c+dcos x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{a+bsin x}{c+dcos x})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(c+dcos x) \frac{d}{dx}(a+bsin x) - (a+bsin x)\frac{d}{dx}(c+dcos x)}{(c+dcos x)^2}\\ f'(x) = \frac{(c+dcos x) [\frac{d}{dx}(a)+\frac{d}{dx}(bsin x)] - (a+bsin x)\frac{d}{dx}(c)+\frac{d}{dx}(dcos x)}{(c+dcos x)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(c+dcos x) [0+b cos x] - (a+bsin x)[0+(-d sin x)]}{(c+dcos x)^2}\\ f'(x) = \frac{(c+dcos x) (b cos x) + (a+bsin x)(d sin x)}{(c+dcos x)^2}\\ f'(x) = \frac{(bc\hspace{0.1cm} cos x+db\hspace{0.1cm} cos^2 x) + (ad\hspace{0.1cm} sin x+db\hspace{0.1cm} sin^2 x)}{(c+d\hspace{0.1cm}cos x)^2}\\ f'(x) = \frac{(bc\hspace{0.1cm} cos x + db (cos^2 x + sin ^2 x) + ad \hspace{0.1cm}sin x)}{(c+d\hspace{0.1cm}cos x)^2}\\ f'(x) = \frac{(bc\hspace{0.1cm} cos x + db+ ad\hspace{0.1cm} sin x)}{(c+d\hspace{0.1cm}cos x)^2}

Question 21: \frac{sin(x+a)}{cos x}

**Solution:

f(x) = \frac{sin(x+a)}{cos x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{sin(x+a)}{cos x})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(cos x) \frac{d}{dx}(sin(x+a)) - (sin(x+a))\frac{d}{dx}(cos x)}{(cos x)^2}\\ f'(x) = \frac{(cos x) \frac{d}{dx}(sin(x+a)) - (sin(x+a))(- sin x)}{(cos x)^2}\\ f'(x) = \frac{(cos x) \frac{d}{dx}(sin(x+a)) + (sin(x+a))(sin x)}{(cos x)^2}

Let's take g(x) = sin (x+a)

g'(x) = \frac{d}{dx}(sin(x+a))

g(x+h) = sin((x+h)+a)

From the first principle,

g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\\ g'(x) = \lim_{h \to 0} (\frac{sin((x+h)+a)-(sin(x+a))}{h})

Using the trigonometric identity,

**sin A - sin B = 2 cos (\frac{A+B}{2}) sin (\frac{A-B}{2})

g'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{x+h+a+(x+a)}{2}) sin (\frac{x+h+a-(x+a)}{2}}{h})\\ g'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{2x+h+2a}{2}) sin (\frac{h}{2})}{h})\\ g'(x) = \lim_{h \to 0} (2 cos (\frac{2x+h+2a}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h}))

Multiply and divide by 2, we have

g'(x) = 2 cos (\frac{2x+0+2a}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) \times \frac{2}{2}\\ g'(x) = 2 cos (x+a) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{\frac{h}{2}})) \times \frac{1}{2}\\ g'(x) = cos (x+a) (1)\\ g'(x) = cos (x+a)

Hence,

f'(x) = \frac{(cos x) \frac{d}{dx}(sin(x+a)) + (sin(x+a))(sin x)}{(cos x)^2}\\ f'(x) = \frac{(cos x) (cos(x+a)) + (sin(x+a))(sin x)}{(cos x)^2}

Using the trigonometric identity,

**cos A cos B + sin A sin B = cos (A-B)

f'(x) = \frac{cos(x+a-x)}{(cos x)^2}\\ f'(x) = \frac{cos(a)}{(cos x)^2}

Question 22: x4(5sin x - 3cos x)

**Solution:

f(x) = x4 (5sin x - 3cos x)

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^4 (5sin x - 3cos x))

Using the product rule, we have

****(uv)' = uv' + vu'**

f'(x) = (x^4) \frac{d}{dx}(5sin x - 3cos x) + (5sin x - 3cos x)\frac{d}{dx}(x^4)

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (x^4) [\frac{d}{dx}(5sin x) - \frac{d}{dx}(3cos x)] + (5sin x - 3cos x)(4x^{4-1})

f'(x) = (x4) [(5 cos x) - (3 (- sin x))] + (5sin x - 3cos x)(4x3)

f'(x) = (x4) [(5 cos x) + (3 sin x)] + (5sin x - 3cos x)(4x3)

f'(x) = (x3) [5x cos x + 3x sin x + 20sin x - 12 cos x]

Question 23: (x2+1) cos x

**Solution:

f(x) = (x2+1) cos x

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}((x^2+1) cos x)

Using the product rule, we have

****(uv)' = uv' + vu'**

f'(x) = (x^2+1) \frac{d}{dx}(cos x) + (cos x)\frac{d}{dx}(x^2+1)

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (x^2+1) (- sin x) + (cos x)[\frac{d}{dx}(x^2)+\frac{d}{dx}(1)]

f'(x) = (x2+1) (- sin x) + (cos x)[(2x2-1)+0]

f'(x) = -x2 sin x- sin x + 2x cos x

Question 24: (ax^2+sin x) (p+q cos x)

**Solution:

f(x) = (ax^2+sin x) (p+q cos x)

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}((ax^2+sin x) (p+q cos x))

Using the product rule, we have

****(uv)' = uv' + vu'**

f'(x) = (ax^2+sin x) \frac{d}{dx}(p+q cos x) + (p+q cos x)\frac{d}{dx}((ax^2+sin x))\\ f'(x) = (ax^2+sin x) [\frac{d}{dx}(p)+\frac{d}{dx}(q cos x)] + (p+q cos x)[\frac{d}{dx}(ax^2)+\frac{d}{dx}(sin x)]

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (ax^2+sin x) [0+q(- sin x)] + (p+q cos x)[a(2x^{2-1})+(cos x)]\\ f'(x) = (ax^2+sin x) (- q sin x) + (p+q cos x)[2ax+cos x]\\ f'(x) = - q sin x(ax^2 + sin x) + (p+q cos x)[2ax+cos x]

Question 25: (x + cos x)(x - tan x)

**Solution:

f(x) = (x + cos x)(x - tan x)

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}((x + cos x)(x - tan x))

Using the product rule, we have

****(uv)' = uv' + vu'**

f'(x) = (x + cos x) \frac{d}{dx}(x - tan x) + (x - tan x)\frac{d}{dx}(x + cos x)\\ f'(x) = (x + cos x) [\frac{d}{dx}(x) - \frac{d}{dx}(tan x)] + (x - tan x)[\frac{d}{dx}(x) + \frac{d}{dx}(cos x)]

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (x + cos x) [1 - \frac{d}{dx}(tan x)] + (x - tan x)[1 + (- sin x)]

Let's take g(x) = tan x

g'(x) = \frac{d}{dx}(tan x)\\ g(x) = tan x = \frac{sin \hspace{0.1cm}x}{cos \hspace{0.1cm}x}\\ g(x+h) = \frac{sin (x+h)}{cos (x+h)}

From the first principle,

g'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h)}{cos (x+h)}-\frac{sin\hspace{0.1cm} x}{cos\hspace{0.1cm} x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{cos \hspace{0.1cm}x \hspace{0.1cm}sin (x+h)-sin\hspace{0.1cm} x \hspace{0.1cm}cos(x+h)}{cos (x+h)cos \hspace{0.1cm}x}}{h}

Using the trigonometric identity,

**sin a cos b - cos a sin b = sin (a-b)

g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h -x)}{cos (x+h)cos x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{sin (h)}{h \hspace{0.1cm}cos (x+h)\hspace{0.1cm}cos\hspace{0.1cm} x}\\ g'(x) = \frac{1}{cos\hspace{0.1cm} x} (\lim_{h \to 0} \frac{1}{cos(x+h)}) (\lim_{h \to 0} \frac{sin h}{h})\\ g'(x) = \frac{1}{cos \hspace{0.1cm}x} \frac{1}{cos(x+0)} (1)\\ g'(x) = \frac{1}{cos^2 x}

g'(x) = sec2x

Hence,

f'(x) = (x + cos x) [1 - \frac{d}{dx}(tan x)] + (x - tan x)[1 + (- sin x)]

f'(x) = (x + cos x) [1 - (sec2 x)] + (x - tan x)[1 - sin x]

f'(x) = (x + cos x) [tan2 x] + (x - tan x)[1 - sin x]

f'(x) = tan2 x(x + cos x) + (x - tan x)[1 - sin x]

Question 26: \frac{4x+5sin x}{3x+7cos x}

**Solution:

f(x) = \frac{4x+5sin x}{3x+7cos x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{4x+5sin x}{3x+7cos x})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(3x+7cos x) \frac{d}{dx}(4x+5sin x) - (4x+5sin x)\frac{d}{dx}(3x+7cos x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(3x+7cos x) [\frac{d}{dx}(4x)+\frac{d}{dx}(5sin x)] - (4x+5sin x)[\frac{d}{dx}(3x)+\frac{d}{dx}(7cos x)]}{(3x+7cos x)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(3x+7cos x) [4+(5 cos x)] - (4x+5sin x)[3 + 7(- sin x)]}{(3x+7cos x)^2}\\ f'(x) = \frac{(3x+7cos x) [4+5 cos x] - (4x+5sin x)[3 - 7 sin x]}{(3x+7cos x)^2}\\ f'(x) = \frac{(12x+28 cos x+15x cos x + 35 cos^2x) - [(12x + 15sin x)-(28x sin x + 35 sin^2 x)]}{(3x+7cos x)^2}\\ f'(x) = \frac{(12x+28 cos x+15x cos x + 35 cos^2x)- (12x + 15sin x-28x sin x - 35 sin^2 x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(12x+28 cos x+15x cos x + 35 cos^2x - 12x - 15sin x+28x sin x + 35 sin^2 x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(28 cos x+15x cos x + 35 (cos^2x + sin^2x) - 15sin x + 28x sin x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(28 cos x+15x cos x + 35 - 15sin x + 28x sin x)}{(3x+7cos x)^2}

Question 27: \frac{x^2 cos(\frac{\pi}{4})}{sin x}

**Solution:

f(x) = \frac{x^2 cos(\frac{\pi}{4})}{sin x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x^2 cos(\frac{\pi}{4})}{sin x})\\ \frac{d}{dx}(f(x)) = cos(\frac{\pi}{4}) \frac{d}{dx}(\frac{x^2)}{sin x})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = (cos(\frac{\pi}{4}) [\frac{(sin x) \frac{d}{dx}(x^2) - (x^2)\frac{d}{dx}(sin x)}{(sin x)^2}]

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = cos(\frac{\pi}{4}) [\frac{(sin x) (2x^{2-1}) - (x^2) (cos x)}{(sin x)^2}]\\ f'(x) = cos(\frac{\pi}{4}) [\frac{2x sin x - x^2 cos x)}{(sin x)^2}]\\ f'(x) = [\frac{(x cos(\frac{\pi}{4})(2 sin x - x cos x)}{(sin x)^2}]

Question 28: \frac{x}{1+tan x}

**Solution:

f(x) = \frac{x}{1+tan x}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x}{1+tan x)})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(1+tan x) \frac{d}{dx}(x) - (x)\frac{d}{dx}(1+tan x)}{(1+tan x)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(1+tan x) (1) - (x)[\frac{d}{dx}(1)+\frac{d}{dx}(tan x)]}{(1+tan x)^2}\\ f'(x) = \frac{(1+tan x) - (x)[0+\frac{d}{dx}(tan x)]}{(1+tan x)^2}

Let's take g(x) = tan x

g'(x) = \frac{d}{dx}(tan x)\\ g(x) = tan x = \frac{sin \hspace{0.1cm}x}{cos \hspace{0.1cm}x}\\ g(x+h) = \frac{sin (x+h)}{cos (x+h)}

From the first principle,

g'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h)}{cos (x+h)}-\frac{sin\hspace{0.1cm} x}{cos\hspace{0.1cm} x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{cos \hspace{0.1cm}x \hspace{0.1cm}sin (x+h)-sin\hspace{0.1cm} x \hspace{0.1cm}cos(x+h)}{cos (x+h)cos \hspace{0.1cm}x}}{h}

Using the trigonometric identity,

**sin a cos b - cos a sin b = sin (a-b)

g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h -x)}{cos (x+h)cos x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{sin (h)}{h \hspace{0.1cm}cos (x+h)\hspace{0.1cm}cos\hspace{0.1cm} x}\\ g'(x) = \frac{1}{cos\hspace{0.1cm} x} (\lim_{h \to 0} \frac{1}{cos(x+h)}) (\lim_{h \to 0} \frac{sin h}{h})\\ g'(x) = \frac{1}{cos \hspace{0.1cm}x} \frac{1}{cos(x+0)} (1)\\ g'(x) = \frac{1}{cos^2 x}

g'(x) = sec2x

Hence, f'(x) = \frac{(1+tan x) - (x)[0+\frac{d}{dx}(tan x)]}{(1+tan x)^2}\\ f'(x) = \frac{1+tan x - x sec^2x}{(1+tan x)^2}

Question 29: (x + sec x) (x-tan x)

**Solution:

f(x) = (x + sec x) (x-tan x)

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}((x + sec x) (x-tan x))

Using the product rule, we have

****(uv)' = uv' + vu'**

f'(x) = (x + sec x) \frac{d}{dx}(x - tan x) + (x - tan x)\frac{d}{dx}(x + sec x)\\ f'(x) = (x + sec x) [\frac{d}{dx}(x) - \frac{d}{dx}(tan x)] + (x - tan x)[\frac{d}{dx}(x) + \frac{d}{dx}(sec x)]

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (x + sec x) [1 - \frac{d}{dx}(tan x)] + (x - tan x)[1 + \frac{d}{dx}(sec x)]

Let's take g(x) = tan x

g'(x) = \frac{d}{dx}(tan x)\\ g(x) = tan x = \frac{sin \hspace{0.1cm}x}{cos \hspace{0.1cm}x}\\ g(x+h) = \frac{sin (x+h)}{cos (x+h)}

From the first principle,

g'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h)}{cos (x+h)}-\frac{sin\hspace{0.1cm} x}{cos\hspace{0.1cm} x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{cos \hspace{0.1cm}x \hspace{0.1cm}sin (x+h)-sin\hspace{0.1cm} x \hspace{0.1cm}cos(x+h)}{cos (x+h)cos \hspace{0.1cm}x}}{h}

Using the trigonometric identity,

**sin a cos b - cos a sin b = sin (a-b)

g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h -x)}{cos (x+h)cos x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{sin (h)}{h \hspace{0.1cm}cos (x+h)\hspace{0.1cm}cos\hspace{0.1cm} x}\\ g'(x) = \frac{1}{cos\hspace{0.1cm} x} (\lim_{h \to 0} \frac{1}{cos(x+h)}) (\lim_{h \to 0} \frac{sin h}{h})\\ g'(x) = \frac{1}{cos \hspace{0.1cm}x} \frac{1}{cos(x+0)} (1)\\ g'(x) = \frac{1}{cos^2 x}

g'(x) = sec2x

Now, let's take h(x) = sec x = \frac{1}{cos x}

h(x+h) = \frac{1}{cos (x+h)}

From the first principle,

h'(x) = \lim_{h \to 0} \frac{h(x+h)-h(x)}{h}\\ h'(x) = \lim_{h \to 0} \frac{\frac{1}{cos (x+h)}-\frac{1}{cos x}}{h}\\ h'(x) = \lim_{h \to 0} \frac{\frac{cos x-cos (x+h)}{cos (x+h)cos x}}{h}\\ h'(x) = \frac{1}{cos x}\lim_{h \to 0} \frac{\frac{cos x-cos (x+h)}{cos (x+h)}}{h}

Using the trigonometric identity,

**cos a - cos b = -2 sin (\frac{a+b}{2}) sin (\frac{a-b}{2})

h'(x) = \frac{1}{cos x}\lim_{h \to 0} \frac{\frac{-2 sin (\frac{x+x+h}{2}) sin (\frac{x-(x+h)}{2})}{cos (x+h)}}{h}\\ h'(x) = \frac{1}{cos x}\lim_{h \to 0} \frac{-2 sin (\frac{2x+h}{2}) sin (\frac{-h}{2})}{hcos (x+h)}\\ h'(x) = \frac{2}{cos x}\lim_{h \to 0} \frac{sin (\frac{2x+h}{2}) sin (\frac{h}{2})}{hcos (x+h)}\\ h'(x) = \frac{2}{cos x}\lim_{h \to 0} \frac{sin (\frac{2x+h}{2})}{cos (x+h)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{h}

Multiply and divide by 2, we have

h'(x) = \frac{2}{cos x}\lim_{h \to 0} \frac{sin (\frac{2x+h}{2})}{cos (x+h)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{h} \times \frac{2}{2}\\ h'(x) = \frac{2}{cos x} \frac{sin (\frac{2x+0}{2})}{cos (x+0)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2}\\ h'(x) = \frac{1}{cos x}(\frac{sin (x)}{cos (x)}) \lim_{h \to 0} \frac{sin (\frac{h}{2})}{\frac{h}{2}}\\ h'(x) = \frac{tan x}{cos x}(1) \\ h'(x) = tan x \hspace{0.1cm}sec x

Hence, f'(x) = (x + sec x) [1 - \frac{d}{dx}(tan x)] + (x - tan x)[1 + \frac{d}{dx}(sec x)]\\ f'(x) = (x + sec x) [1 - (sec^2x)] + (x - tan x)[1 + (sec x \hspace{0.1cm}tan x)]\\ f'(x) = (x + sec x) [tan^2x)] + (x - tan x)[1 + (sec x \hspace{0.1cm}tan x)]

Question 30: \frac{x}{sin^nx}

**Solution:

f(x) = \frac{x}{sin^nx}

Taking derivative both sides,

\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x}{sin^nx})

Using the quotient rule, we have

(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(sin^nx) \frac{d}{dx}(x) - (x)\frac{d}{dx}(sin^nx)}{(sin^nx)^2}

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = \frac{(sin^nx) (1) - (x)\frac{d}{dx}(sin^nx)}{(sin^nx)^2}

Let's take, g(x) = sinn x

When n = 1,

g(x) = sin x

g'(x) = \frac{d}{dx}(sin x) = cos x

When n = 2,

g(x) = sin^2 x = sin x sin x\\ g'(x) = \frac{d}{dx}(sin x\hspace{0.1cm} sin x)

Using the product rule, we have

****(uv)' = uv'+vu'**

g'(x) = (sin x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin x)

g'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

g(x) = sin3 x = sin2 x sin x

g'(x) = \frac{d}{dx}(sin^2 x\hspace{0.1cm} sin x)

Using the product rule, we have

****(uv)' = uv'+vu'**

g'(x) = (sin^2 x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin^2 x)\\ g'(x) = (sin^2 x) (cos x) + (sin x) (2 sin x \hspace{0.1cm}cos x)\\ g'(x) = (sin^2 x \hspace{0.1cm}cos x) + (2 sin^2 x\hspace{0.1cm} cos x)\\ g'(x) = (sin^2 x\hspace{0.1cm} cos x)[1+2]\\ g'(x) = 3 sin^2 x\hspace{0.1cm} cos x

Pattern w.r.t n is seen here, as follows

\frac{d}{dx}(sin^n x) = n sin^{n-1}x\hspace{0.1cm} cos x

Let's check this statement.

For P(n) = n sin^{n-1}x \hspace{0.1cm}cos x

For P(1),

P(1) = 1 sin^{1-1}x \hspace{0.1cm}cos x = cos x . Which is true.

n=k

P(k) = k sin^{k-1}x \hspace{0.1cm}cos x

n = k+1

P(k+1) = \frac{d}{dx}(sin^{k+1} x) = \frac{d}{dx}(sin^k x \hspace{0.1cm}sin x)

Using the product rule, we have

****(uv)' = uv'+vu'**

= (sin^k x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin^k x)\\ = (sin^k x) (cos x) + (sin x) (k sin^{k-1}x \hspace{0.1cm}cos x)\\ = (sin^k x) (cos x)[k+1]

Hence proved for P(k+1).

So, \frac{d}{dx}(sin^n x) = n sin^{n-1}x \hspace{0.1cm}cos x is true.

So, the given equation will be

f'(x) = \frac{(sin^nx) (1) - (x)\frac{d}{dx}(sin^nx)}{sin^{2n}x}\\ f'(x) = \frac{(sin^nx) - x(n sin^{n-1}x cos x)}{sin^{2n}x}