Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.5 | Set 1 (original) (raw)

Last Updated : 23 Aug, 2024

Question 1. Differentiate y = x1/x with respect to x.

**Solution:

We have,

=> y = x1/x

On taking log of both the sides, we get,

=> log y = log x1/x

=> log y = (1/x) (log x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(\frac{logx}{x})

=> \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}(\frac{1}{x})+logx(\frac{-1}{x^2})

=> \frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}-\frac{logx}{x^2}

=> \frac{1}{y}\frac{dy}{dx}=\frac{1-logx}{x^2}

=> \frac{dy}{dx}=\frac{(1-logx)y}{x^2}

=> \frac{dy}{dx}=\frac{(1-logx)x^{\frac{1}{x}}}{x^2}

Question 2. Differentiate y = xsin x with respect to x.

**Solution:

We have,

=> y = xsin x

On taking log of both the sides, we get,

=> log y = log xsin x

=> log y = sin x log x

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(sinxlogx)

=> \frac{1}{y}\frac{dy}{dx}=sinx(\frac{1}{x})+logx(cosx)

=> \frac{1}{y}\frac{dy}{dx}=\frac{sinx}{x}+logxcosx

=> \frac{dy}{dx}=y\left[\frac{sinx}{x}+logxcosx\right]

=> \frac{dy}{dx}=x^{sinx}\left[\frac{sinx}{x}+logxcosx\right]

Question 3. Differentiate y = (1 + cos x)x with respect to x.

**Solution:

We have,

=> y = (1 + cos x)x

On taking log of both the sides, we get,

=> log y = log (1 + cos x)x

=> log y = x log (1 + cos x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (1 + cos x)]

=> \frac{1}{y}\frac{dy}{dx}=-xsinx(\frac{1}{1+cosx})+log(1+cosx)(1)

=> \frac{1}{y}\frac{dy}{dx}=-xsinx(\frac{1}{1+cosx})+log(1+cosx)(1)

=> \frac{1}{y}\frac{dy}{dx}=\frac{-xsinx}{1+cosx}+log(1+cosx)

=> \frac{dy}{dx}=y\left[log(1+cosx)-\frac{xsinx}{1+cosx}\right]

=> \frac{dy}{dx}=(1+cos x)^x\left[log(1+cosx)-\frac{xsinx}{1+cosx}\right]

Question 4. Differentiate y=x^{cos^{-1}x} with respect to x.

**Solution:

We have,

=> y=x^{cos^{-1}x}

On taking log of both the sides, we get,

=> log y = log x^{cos^{-1}x}

=> log y = cos−1 x log x

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(cos^{−1} x log x)

=> \frac{1}{y}\frac{dy}{dx}=cos^{-1}x(\frac{1}{x})+logx(\frac{-1}{\sqrt{1-x^2}})

=> \frac{1}{y}\frac{dy}{dx}=\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}

=> \frac{dy}{dx}=y\left[\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}\right]

=> \frac{dy}{dx}=x^{cos^{-1}x}\left[\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}\right]

Question 5. Differentiate y = (log x)x with respect to x.

**Solution:

We have,

=> y = (log x)x

On taking log of both the sides, we get,

=> log y = log (log x)x

=> log y = x log (log x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (log x)]

=> \frac{1}{y}\frac{dy}{dx}=x(\frac{1}{logx})(\frac{1}{x})+log(logx)

=> \frac{1}{y}\frac{dy}{dx}=\frac{1}{logx}+log(logx)

=> \frac{dy}{dx}=y\left[\frac{1}{logx}+log(logx)\right]

=> \frac{dy}{dx}=(logx)^x\left[\frac{1}{logx}+log(logx)\right]

Question 6. Differentiate y = (log x)cos x with respect to x.

**Solution:

We have,

=> y = (log x)cos x

On taking log of both the sides, we get,

=> log y = log (log x)cos x

=> log y = cos x log (log x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[cos x log (log x)]

=> \frac{1}{y}\frac{dy}{dx}=cosx(\frac{1}{logx})(\frac{1}{x})+log(logx)(-sinx)

=> \frac{1}{y}\frac{dy}{dx}=\frac{cosx}{xlogx}-sinxlog(logx)

=> \frac{dy}{dx}=y\left[\frac{cosx}{xlogx}-sinxlog(logx)\right]

=> \frac{dy}{dx}=(logx)^{cosx}\left[\frac{cosx}{xlogx}-sinxlog(logx)\right]

Question 7. Differentiate y = (sin x)cos x with respect to x.

**Solution:

We have,

=> y = (sin x)cos x

On taking log of both the sides, we get,

=> log y = log (sin x)cos x

=> log y = cos x log (sin x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[cos x log (sin x)]

=> \frac{1}{y}\frac{dy}{dx}=cosx(\frac{1}{sinx})(cosx)+log(sinx)(-sinx)

=> \frac{1}{y}\frac{dy}{dx}=\frac{cos^2x}{sinx}-sinxlog(sinx)

=> \frac{1}{y}\frac{dy}{dx}=cotxcosx-sinxlog(sinx)

=> \frac{dy}{dx}=y\left[cotxcosx-sinxlog(sinx)\right]

=> \frac{dy}{dx}=(sinx)^{cosx}\left[cotxcosx-sinxlog(sinx)\right]

Question 8. Differentiate y = ex log x with respect to x.

**Solution:

We have,

=> y=ex log x

=> y = e^{logx^x}

=> y = xx

On taking log of both the sides, we get,

=> log y = log xx

=> log y = x log x

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x log x)

=> \frac{1}{y}\frac{dy}{dx}=x(\frac{1}{x})+logx

=> \frac{1}{y}\frac{dy}{dx}=1+logx

=> \frac{dy}{dx}=y(1+logx)

=> \frac{dy}{dx}=x^{x}(1+logx)

Question 9. Differentiate y = (sin x)log x with respect to x.

**Solution:

We have,

=> y = (sin x)log x

On taking log of both the sides, we get,

=> log y = log (sin x)log x

=> log y = log x log (sin x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log x log (sin x)]

=> \frac{1}{y}\frac{dy}{dx}=logx(\frac{1}{sinx})(cosx)+log(sinx)(\frac{1}{x})

=> \frac{1}{y}\frac{dy}{dx}=logxcotx+\frac{log(sinx)}{x}

=> \frac{dy}{dx}=y\left[logxcotx+\frac{log(sinx)}{x}\right]

=> \frac{dy}{dx}=(sinx)^{logx}\left[logxcotx+\frac{log(sinx)}{x}\right]

Question 10. Differentiate y = 10log sin x with respect to x.

**Solution:

We have,

=> y = 10log sin x

On taking log of both the sides, we get,

=> log y = log 10log sin x

=> log y = log (sin x) log 10

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log (sin x) log 10]

=> \frac{1}{y}\frac{dy}{dx}=log10\frac{d}{dx}[log(sinx)]

=> \frac{1}{y}\frac{dy}{dx}=log10(\frac{1}{sinx})(cosx)

=> \frac{1}{y}\frac{dy}{dx}=log10cotx

=> \frac{dy}{dx}=ylog10cotx

=> \frac{dy}{dx}=10^{logsinx}[log10cotx]

Question 11. Differentiate y = (log x)log x with respect to x.

**Solution:

We have,

=> y = (log x)log x

On taking log of both the sides, we get,

=> log y = log (log x)log x

=> log y = log x log (log x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log x log (log x)]

=> \frac{1}{y}\frac{dy}{dx}=logx(\frac{1}{logx})(\frac{1}{x})+log(logx)(\frac{1}{x})

=> \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{log(logx)}{x}

=> \frac{1}{y}\frac{dy}{dx}=\frac{1+log(logx)}{x}

=> \frac{dy}{dx}=\frac{y[1+log(logx)]}{x}

=> \frac{dy}{dx}=\frac{(logx)^{logx}[1+log(logx)]}{x}

Question 12. Differentiate y = 10^{10^x} with respect to x.

**Solution:

We have,

=> y = 10^{10^x}

On taking log of both the sides, we get,

=> log y = log 10^{10^x}

=> log y = 10x log 10

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(10^x log 10)

=> \frac{1}{y}\frac{dy}{dx}=log10(10^xlog10)

=> \frac{1}{y}\frac{dy}{dx}=10^x(log10)^2

=> \frac{dy}{dx}=y10^x(log10)^2

=> \frac{dy}{dx}=10^{10^x}\left[10^x(log10)^2\right]

=> \frac{dy}{dx}=(10^{10^x+x})(log10)^2

Question 13. Differentiate y = sin xx with respect to x.

**Solution:

We have,

=> y = sin xx

=> sin−1 y = xx

On taking log of both the sides, we get,

=> log (sin−1 y) = log xx

=> log (sin−1 y) = x log x

On differentiating both sides with respect to x, we get,

=> (\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=\frac{d}{dx}(xlogx)

=> (\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=x(\frac{1}{x})+logx

=> (\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=1+logx

=> \frac{dy}{dx}=(1+logx)(sin^{-1}y)(\sqrt{1-y^2})

=> \frac{dy}{dx}=(1+logx)(sin^{-1}(sinx^x))(\sqrt{1-(sinx^x)^2})

=> \frac{dy}{dx}=(1+logx)(x^x)(\sqrt{1-sin^2x^x})

=> \frac{dy}{dx}=(1+logx)(x^x)(\sqrt{cos^2x^x})

=> \frac{dy}{dx}=x^xcosx^x(1+logx)

Question 14. Differentiate y = (sin−1x)x with respect to x.

**Solution:

We have,

=> y = (sin−1x)x

On taking log of both the sides, we get,

=> log y = (sin−1x)x

=> log y = x log (sin−1x)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (sin^{−1}x)]

=> \frac{1}{y}\frac{dy}{dx}=x(\frac{1}{sin^{-1}x})(\frac{1}{\sqrt{1-x^2}})+log (sin^{−1}x)

=> \frac{1}{y}\frac{dy}{dx}=\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)

=> \frac{dy}{dx}=y\left[\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)\right]

=> \frac{dy}{dx}=(sin^{-1}x)^x\left[\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)\right]

Question 15. Differentiate y=x^{sin^{-1}x} with respect to x.

**Solution:

We have,

=> y=x^{sin^{-1}x}

On taking log of both the sides, we get,

=> log y = log x^{sin^{-1}x}

=> log y = sin−1x log x

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(sin^{−1}x log x)

=> \frac{1}{y}\frac{dy}{dx}=sin^{−1}x(\frac{1}{x})+logx(\frac{1}{\sqrt{1-x^2}})

=> \frac{1}{y}\frac{dy}{dx}=\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}

=> \frac{dy}{dx}=y\left[\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}\right]

=> \frac{dy}{dx}=x^{sin^{-1}x}\left[\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}\right]

Question 16. Differentiate y=(tanx)^{\frac{1}{x}} with respect to x.

**Solution:

We have,

=> y=(tanx)^{\frac{1}{x}}

On taking log of both the sides, we get,

=> log y = log (tanx)^{\frac{1}{x}}

=> log y = \frac{1}{x}log(tanx)

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[\frac{1}{x}log(tanx)]

=> \frac{1}{y}\frac{dy}{dx}=(\frac{1}{x})(\frac{1}{tanx})(sec^2x)+log(tanx)(\frac{-1}{x^2})

=> \frac{1}{y}\frac{dy}{dx}=\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}

=> \frac{dy}{dx}=y\left[\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}\right]

=> \frac{dy}{dx}=(tanx)^{\frac{1}{x}}\left[\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}\right]

Question 17. Differentiate y=x^{tan^{-1}x} with respect to x.

**Solution:

We have,

=> y=x^{tan^{-1}x}

On taking log of both the sides, we get,

=> log y = log y=x^{tan^{-1}x}

=> log y = tan−1 x log x

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(tan^{−1} x log x)

=> \frac{1}{y}\frac{dy}{dx}=tan^{−1}x(\frac{1}{x})+logx(\frac{1}{1+x^2})

=> \frac{1}{y}\frac{dy}{dx}=\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}

=> \frac{dy}{dx}=y\left[\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}\right]

=> \frac{dy}{dx}=x^{tan^{-1}x}\left[\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}\right]

Question 18. Differentiate the following with respect to x.

(i) y = xx √x

**Solution:

We have,

=> y = xx √x

On taking log of both the sides, we get,

=> log y = log (xx √x)

=> log y = log xx + log √x

=> log y = x log x + \frac{1}{2}logx

On differentiating both sides with respect to x, we get,

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x log x +\frac{1}{2}logx)

=> \frac{1}{y}\frac{dy}{dx}=x(\frac{1}{x})+logx+(\frac{1}{2})(\frac{1}{x})

=> \frac{1}{y}\frac{dy}{dx}=1+logx+\frac{1}{2x}

=> \frac{dy}{dx}=y\left(1+logx+\frac{1}{2x}\right)

=> \frac{dy}{dx}=x^x\sqrt{x}\left(1+logx+\frac{1}{2x}\right)

(ii) y=x^{sinx-cosx}+\frac{x^2-1}{x^2+1}

**Solution:

We have,

=> y=x^{sinx-cosx}+\frac{x^2-1}{x^2+1}

=> y=e^{logx^{sinx-cosx}}+\frac{x^2-1}{x^2+1}

=> y=e^{(sinx-cosx)logx}+\frac{x^2-1}{x^2+1}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=\frac{d}{dx}(e^{(sinx-cosx)logx}+\frac{x^2-1}{x^2+1})

=> \frac{dy}{dx}=(e^{(sinx-cosx)logx})[(sinx-cosx)\frac{1}{x}+logx(cosx+sinx)]+\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}

=> \frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{2x(x^2+1-x^2+1)}{(x^2+1)^2}

=> \frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{2x(2)}{(x^2+1)^2}

=> \frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{4x}{(x^2+1)^2}

(iii) y=x^{xcosx}+\frac{x^2+1}{x^2-1}

**Solution:

We have,

=> y=x^{xcosx}+\frac{x^2+1}{x^2-1}

=> y=e^{logx^{xcosx}}+\frac{x^2+1}{x^2-1}

=> y=e^{xcosxlogx}+\frac{x^2+1}{x^2-1}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=(e^{xcosxlogx})[x((-sinx)logx+cosx(\frac{1}{x}))+cosxlogx]+\frac{(x^2-1)(2x)-(x^2+1)(2x)}{(x^2-1)^2}

=> \frac{dy}{dx}=(e^{xcosxlogx})[-xsinxlogx+cosx+cosxlogx]+\frac{2x(x^2-1-x^2-1)}{(x^2-1)^2}

=> \frac{dy}{dx}=(e^{xcosxlogx})[-xsinxlogx+cosx(1+logx)]+\frac{2x(-2)}{(x^2-1)^2}

=> \frac{dy}{dx}=(e^{xcosxlogx})[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}

=> \frac{dy}{dx}=x^{xcosx}[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}

(iv) y = (x cos x)x + (x sin x)1/x

**Solution:

We have,

=> y=(x cos x)x + (x sin x)1/x

=> y=e^{log(x cos x)^x}+ e^{log(x sin x)^{\frac{1}{x}}}

=> y=e^{xlog(x cos x)}+ e^{\frac{1}{x}log(x sin x)}

=> y=e^{x(logx+logcosx)}+ e^{\frac{1}{x}(logx+logsin x)}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=(e^{x(logx+logcosx)})[x(\frac{1}{x}+(\frac{1}{cosx})(-sinx))+log(xcosx)(1)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1}{x}(\frac{1}{x}+(\frac{1}{sinx})(cosx))+log(xsinx)(\frac{-1}{x^2})]

=> \frac{dy}{dx}=(e^{x(logx+logcosx)})[1-xtanx+log(xcosx)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1}{x^2}+\frac{1}{xcotx}-\frac{log(xsinx)}{x^2}]

=> \frac{dy}{dx}=(e^{x(logx+logcosx)})[1-xtanx+log(xcosx)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1-log(xsinx)+xcotx}{x^2}]

=> \frac{dy}{dx}=(xcosx)^x[1-xtanx+log(xcosx)]+ (xsinx)^{\frac{1}{x}}[\frac{1-log(xsinx)+xcotx}{x^2}]

(v) y=(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}

**Solution:

We have,

=> y=(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}

=> y=e^{log(x+\frac{1}{x})^x}+e^{logx^{(1+\frac{1}{x})}}

=> y=e^{xlog(x+\frac{1}{x})}+e^{(1+\frac{1}{x})logx}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=(e^{xlog(x+\frac{1}{x})})[x(\frac{1}{x+\frac{1}{x}})(1-\frac{1}{x^2})+log(x+\frac{1}{x})]+(e^{(1+\frac{1}{x})logx})[(1+\frac{1}{x})(\frac{1}{x})+logx(\frac{-1}{x^2})]

=> \frac{dy}{dx}=(e^{xlog(x+\frac{1}{x})})[(\frac{x-\frac{1}{x}}{x+\frac{1}{x}})+log(x+\frac{1}{x})]+(e^{(1+\frac{1}{x})logx})[\frac{1}{x}+\frac{1}{x^2}-logx(\frac{1}{x^2})]

=> \frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}[\frac{1}{x}+\frac{1}{x^2}-logx(\frac{1}{x^2})]

=> \frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}[\frac{1}{x}+\frac{1}{x^2}-\frac{logx}{x^2}]

=> \frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}(\frac{x+1-logx}{x^2})

(vi) y = esin x + (tan x)x

**Solution:

We have,

=> y = esin x + (tan x)x

=> y = e^{sin x} + e^{log(tanx)^x}

=> y = e^{sin x} + e^{xlog(tanx)}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=\frac{d}{dx}(e^{sin x} + e^{xlog(tanx)})

=> \frac{dy}{dx}=(e^{sin x})(cosx) + (e^{xlogtanx})\left[x(\frac{1}{tanx})(sec^2x)+log(tanx)\right]

=> \frac{dy}{dx}=e^{sin x}cosx+(tanx)^x\left[\frac{xsec^2x}{tanx}+log(tanx)\right]

(vii) y = (cos x)x + (sin x)1/x

**Solution:

We have,

=> y = (cos x)x + (sin x)1/x

=> y = e^{log(cos x)^x} + e^{log(sin x)^{1/x}}

=> y = e^{xlog(cos x)} + e^{\frac{1}{x}log(sin x)}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=(e^{xlog(cos x)})[x(\frac{1}{cosx})(-sinx)+log(cosx)] + (e^{\frac{1}{x}log(sin x)})[\frac{1}{x}(\frac{1}{sinx}(cosx))+log(sinx)(-\frac{1}{x^2})]

=> \frac{dy}{dx}=(e^{xlog(cos x)})[-xtanx+log(cosx)] + (e^{\frac{1}{x}log(sin x)})[\frac{cotx}{x}-\frac{log(sinx)}{x^2}]

=> \frac{dy}{dx}=(cosx)^x[-xtanx+log(cosx)] + (sinx)^{\frac{1}{x}}[\frac{cotx}{x}-\frac{log(sinx)}{x^2}]

****(**viii) y=x^{x^2-3}+(x-3)^{x^2} , for x > 3

**Solution:

We have,

=> y=x^{x^2-3}+(x-3)^{x^2}

=> y=e^{logx^{x^2-3}}+e^{log(x-3)^{x^2}}

=> y=e^{(x^2-3)logx}+e^{x^2log(x-3)}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=(e^{(x^2-3)logx})[(x^2-3)(\frac{1}{x})+logx(2x)]+(e^{x^2log(x-3)})[x^2(\frac{1}{x-3})+log(x-3)(2x)]

=> \frac{dy}{dx}=(e^{(x^2-3)logx})[\frac{x^2-3}{x}+2xlogx]+(e^{x^2log(x-3)})[\frac{x^2}{x-3}+2xlog(x-3)]

=> \frac{dy}{dx}=x^{x^2-3}[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]

Question 19. Find dy/dx when y = ex + 10x + xx.

**Solution:

We have,

=> y = ex + 10x + xx

=> y = e^x + 10^x + e^{logx^x}

=> y = e^x + 10^x + e^{xlogx}

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=\frac{d}{dx}(e^x + 10^x + e^{xlogx})

=> \frac{dy}{dx}=e^x+10^xlog10+e^{xlogx}[x(\frac{1}{x})+logx]

=> \frac{dy}{dx}=e^x+10^xlog10+x^x(1+logx)

=> \frac{dy}{dx}=e^x+10^xlog10+x^x(logex)

Question 20. Find dy/dx when y = xn + nx + xx + nn.

**Solution:

We have,

=> y = xn + nx + xx + nn

=> y=x^n + n^x + e^{logx^x} + n^n

=> y=x^n + n^x + e^{xlogx} + n^n

On differentiating both sides with respect to x, we get,

=> \frac{dy}{dx}=\frac{d}{dx}(x^n+n^x+e^{xlogx}+n^n)

=> \frac{dy}{dx}=nx^{n-1}+n^xlogn+e^{xlogx}[x(\frac{1}{x})+logx]+0

=> \frac{dy}{dx}=nx^{n-1}+n^xlogn+x^x(1+logx)

=> \frac{dy}{dx}=nx^{n-1}+n^xlogn+x^x(logex)

Summary

Exercise 11.5 in RD Sharma's Class 12 Mathematics textbook focuses on the application of derivatives to find the maximum and minimum values of functions. This set covers problems involving optimization, where students must determine the extreme values of functions within given intervals or under specific constraints. The questions typically require students to find critical points, analyze the behavior of functions, and apply the first and second derivative tests to identify local and global extrema.

Practice Questions

**1. Find the maximum and minimum values of f(x) = x³ - 12x + 4 on the closed interval [-2, 3].

**2. Determine the dimensions of a rectangular box with a square base and a surface area of 150 cm², which has the maximum possible volume.

**3. Find the points on the curve y = x³ - 3x that are nearest to the point (0, 1).

**4. A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 10 meters, find the dimensions that will admit the maximum amount of light.

**5. Find the maximum area of a right-angled triangle that can be inscribed in a circle of radius 5 units.

**6. Determine the minimum value of |x - 2| + |x + 2| for all real values of x.

**7. A piece of wire 20 cm long is cut into two parts. One part is bent into a square and the other into a circle. How should the wire be cut so that the sum of the areas enclosed is a minimum?

**8. Find the points on the parabola y = x² that are closest to the point (3, 0).

**9. A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 feet, find the dimensions of the window that maximize the area.

**10. Find the maximum product of two positive numbers whose sum is 20.