Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.1 (original) (raw)

Last Updated : 23 Aug, 2024

**Question 1. Differentiate the following functions from first principles e-x

**Solution:

We have,

Let,

f(x)=e-x

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

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{e^{-(x+h)}-e^{-x}}{h}

=\lim_{h\to0}\frac{e^{-x}.e^{-h}-e^{-x}}{h}

=\lim_{h\to0}[e^{-x}(\frac{e^{-h}-1}{-h})](-1)

=-e^{-x}[\lim_{h\to0}(\frac{e^{-h}-1}{-h})]

=-e-x

**Question 2. Differentiate the following functions from first principles e3x

**Solution:

We have,

Let,

f(x)=e3x

f(x+h)=e3(x+h)

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{e^{3(x+h)}-e^{3x}}{h}

=\lim_{h\to0}\frac{e^{3x}.e^{3h}-e^{3x}}{h}

=\lim_{h\to0}[e^{3x}(\frac{e^{3h}-1}{3h})](3)

=3e^{3x}[\lim_{h\to0}(\frac{e^{3h}-1}{3h})]

=3e3x

**Question 3. Differentiate the following functions from first principles eax+b

**Solution:

We have,

Let,

f(x)=eax+b

f(x+h)=ea(x+h)+b

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{e^{a(x+h)+b}-e^{ax+b}}{h}

=\lim_{h\to0}\frac{e^{ax+b}.e^{ah}-e^{ax+b}}{h}

=\lim_{h\to0}[e^{ax+b}(\frac{e^{ah}-1}{ah})](a)

=ae^{ax+b}[\lim_{h\to0}(\frac{e^{ah}-1}{ah})]

=aeax+b

**Question 4. Differentiate the following functions from first principles ecosx

**Solution:

We have,

Let,

f(x)=ecosx

f(x+h)=ecos(x+h)

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{e^{cos(x+h)}-e^{cosx}}{h}

=\lim_{h\to0}[e^{cosx}(\frac{e^{cos(x+h)-cosx}-1}{h})]

=e^{cosx}\lim_{h\to0}(\frac{e^{cos(x+h)-cosx}-1}{cos(x+h)-cosx})(\frac{cos(x+h)-cosx}{h})

=e^{cosx}\lim_{h\to0}(\frac{cos(x+h)-cosx}{h})

=e^{cosx}\lim_{h\to0}(\frac{-2sin(\frac{x+h+x}{2})sin(\frac{x+h-x}{2})}{h})

=e^{cosx}\lim_{h\to0}\frac{-2sin(\frac{2x+h}{2})}{2}\frac{sin\frac{h}{2}}{\frac{h}{2}}

=e^{cosx}\lim_{h\to0}\frac{-2sin(\frac{2x+h}{2})}{2}\frac{1}{2}

=ecosx(-sinx)

=-sinx.ecosx

**Question 5. Differentiate the following functions from first principles e√2x

**Solution:

We have,

Let,

f(x)=e√2x

f(x+h)=e√2(x+h)

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}

=\lim_{h\to0}[e^{\sqrt{2x}}(\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{h})]

=e^{\sqrt{2x}}\lim_{h\to0}(\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{\sqrt{2(x+h)}-\sqrt{2x}})(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h})

=e^{\sqrt{2x}}\lim_{h\to0}(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h})

=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2(x+h)-2x}{h(\sqrt{2(x+h)}-\sqrt{2x})}) (After rationalising the numerator)

=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2x+2h-2x}{h(\sqrt{2(x+h)}-\sqrt{2x})})

=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2h}{h(\sqrt{2(x+h)}-\sqrt{2x})})

=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2}{(\sqrt{2(x+h)}-\sqrt{2x})})

=\frac{e^{\sqrt{2x}}}{\sqrt{2x}}

**Question 6. Differentiate the following functions from first principles log(cosx)

**Solution:

We have,

Let,

f(x)=log(cosx)

f(x+h)=log(cos(x+h))

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{log(cos(x+h))-log(cosx)}{h}

=\lim_{h\to0}\frac{log\frac{cos(x+h)}{cosx}}{h}

=\lim_{h\to0}\frac{log[1+\frac{cos(x+h)}{cosx}-1]}{h}

=\lim_{h\to0}\frac{log[1+\frac{cos(x+h)}{cosx}]}{\frac{cos(x+h)-cosx}{cosx}}×\lim_{h\to0}\frac{cos(x+h)-cosx}{cosx}

=1×\lim_{h\to0}\frac{cos(x+h)-cosx}{h×cosx}

=\lim_{h\to0}\frac{-2sin(\frac{x+h+x}{2})sin(\frac{x+h-x}{2})}{h×cosx}

=-2\lim_{h\to0}\frac{sin(\frac{2x+h}{2})sin(\frac{h}{2})}{2(\frac{h}{2})×cosx}

Since, \lim_{h\to0}\frac{sinx}{x}=1

=-(2sinx)/(2cosx)

=-tanx

**Question 7. Differentiate the following functions from first principles e√cotx

**Solution:

We have,

Let,

f(x)=e√cotx

f(x+h)=e√cot(x+h)

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{e^{\sqrt{cot(x+h)}}-e^{\sqrt{cotx}}}{h}

=\lim_{h\to0}[e^{\sqrt{cotx}}(\frac{e^{\sqrt{cot(x+h)}-\sqrt{cotx}}-1}{h})]

=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{e^{\sqrt{cot(x+h)}-\sqrt{cotx}}-1}{\sqrt{cot(x+h)}-\sqrt{cotx}})(\frac{\sqrt{cot(x+h)}-\sqrt{cotx}}{h})

=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{\sqrt{cot(x+h)}-\sqrt{cotx}}{h})

since, \lim_{h\to0}\frac{e^x-1}{x}=1

=e^{\sqrt{cotx}}\lim_{h\to0}{\frac{cot(x+h)-cotx}{h(\sqrt{cot(x+h)}-\sqrt{cotx})}} (After rationalising the numerator)

=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{\frac{cot(x+h)cotx+1}{cot(x-x-h)}}{h(\sqrt{cot(x+h)}-\sqrt{cotx})})

=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{cot(x+h)cotx+1}{hcot(-h)(\sqrt{cot(x+h)}-\sqrt{cotx})})

=-e^{\sqrt{cotx}}\lim_{h\to0}(\frac{cot(x+h)cotx+1}{\frac{h}{tanh}(\sqrt{cot(x+h)}-\sqrt{cotx})})

Since, \lim_{h\to0}\frac{tanx}{x}=1

=\frac{e^{\sqrt{cotx}}×(cot^2x+1)}{2\sqrt{cotx}}

=\frac{e^{\sqrt{cotx}}×cosec^2x}{2\sqrt{cotx}}

**Question 8. Differentiate the following functions from first principles x2ex

**Solution:

We have,

Let,

f(x)=x2ex

f(x+h)=(x+h)2e(x+h)

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{(x+h)^2e^{(x+h)}-x^2e^x}{h}

=\lim_{h\to0}(\frac{x^2e^{x+h}-x^2e^x}{h}+\frac{2hxe^{x+h}}{h}+\frac{h^2e^{x+h}}{h})

=\lim_{h\to0}(\frac{x^2e^x(e^h-1)}{h}+2xe^{x+h}+{he^{x+h}})

Since, \frac{e^h-1}{h}=1

=x2ex+2xex+0

=ex(x2+2x)

**Question 9. Differentiate the following functions from first principles log(cosecx)

**Solution:

We have,

Let,

f(x)=log(cosecx)

f(x+h)=log(cosec(x+h))

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{log(cosec(x+h))-log(cosecx)}{h}

=\lim_{h\to0}\frac{log\frac{cosec(x+h)}{cosecx}}{h}

=\lim_{h\to0}\frac{log[1+\frac{cosec(x+h)}{cosecx}-1]}{h}

=\lim_{h\to0}\frac{log[1+\frac{sinx}{sin(x+h)}-1]}{h}

=\lim_{h\to0}\frac{log[1+\frac{sinx-sin(x+h)}{sin(x+h)}]}{\frac{sinx-sin(x+h)}{sin(x+h)}}×\lim_{h\to0}\frac{\frac{sinx-sin(x+h)}{sin(x+h)}}{h}

=\lim_{h\to0}\frac{sinx-sin(x+h)}{sin(x+h)}

=\lim_{h\to0}\frac{2cos(\frac{x+x+h}{2})sin(\frac{x-x-h}{2})}{hsin(x+h)}

=\lim_{h\to0}\frac{2cos(\frac{2x+h}{2})sin(\frac{x-x-h}{2})}{sin(x+h)}\frac{sin(-\frac{h}{2})}{-\frac{h}{2}.(-2)}

=-cotx

**Question 10. Differentiate the following functions from first principles sin-1(2x+3)

**Solution:

We have,

Let,

f(x)=sin-1(2x+3)

f(x+h)=sin-1[2(x+h)+3]

f(x+h)=sin-1(2x+2h+3)

\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

=\lim_{h\to0}\frac{sin^{-1}(2x+2h+3)-sin^{-1}(2x+3)}{h}

=\lim_{h\to0}\frac{sin^{-1}[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]}{h}

=\lim_{h\to0}\frac{sin^{-1}t}{t}\frac{t}{h}

Where t=[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]

=\lim_{h\to0}\frac{t}{h}

=\lim_{h\to0}\frac{[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]}{h}

=\lim_{h\to0}\frac{[(2x+2h+3)^2[1-(2x+3)^2]-(2x+3)^2[1-(2x+2h+3)^2]}{h[(2x+2h+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+2h+3)^2]}} (After rationalising the numerator)

Solving above equation

=\lim_{h\to0}\frac{4h[h+(2x+3)]}{h[(2x+2h+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+2h+3)^2]}}

=\frac{4(2x+3)}{[(2x+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+3)^2]}}

=\frac{4(2x+3)}{2[(2x+3)\sqrt{1-(2x+3)^2}]}

=\frac{2}{2[\sqrt{1-(2x+3)^2}]}

Summary

Exercise 11.1 focuses on applying differentiation techniques to various types of functions. It covers the differentiation of algebraic, trigonometric, exponential, and logarithmic functions. Students are expected to use basic differentiation rules, chain rule, product rule, and quotient rule. The exercise also includes problems involving implicit differentiation and higher-order derivatives.