Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.5 | Set 1

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

Solution:

We have,

=> y = x^1/x

On taking log of both the sides, we get,

=> log y = log x^1/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 = x^{sin x} with respect to x.

Solution:

We have,

=> y = x^{sin x}

On taking log of both the sides, we get,

=> log y = log x^{sin 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 = e^{x log x} with respect to x.

Solution:

We have,

=> y=e^{x log x}

=> y = $e^{logx^x}$

=> y = x^x

On taking log of both the sides, we get,

=> log y = log x^x

=> 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 = 10^{log sin x} with respect to x.

Solution:

We have,

=> y = 10^{log sin x}

On taking log of both the sides, we get,

=> log y = log 10^{log 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 = 10^x 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 x^x with respect to x.

Solution:

We have,

=> y = sin x^x

=> sin⁻¹ y = x^x

On taking log of both the sides, we get,

=> log (sin⁻¹ y) = log x^x

=> 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 = x^x √x

Solution:

We have,

=> y = x^x √x

On taking log of both the sides, we get,

=> log y = log (x^x √x)

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

Solution:

We have,

=> y = e^{sin 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 = e^x + 10^x + x^x.

Solution:

We have,

=> y = e^x + 10^x + x^x

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

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

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 + n^x+ x^x + nⁿ.

Solution:

We have,

=> y = xⁿ + n^x + x^x + 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,

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

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

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

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

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Last Updated : 26 May, 2021

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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.5 | Set 1

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

Question 2. Differentiate y = x^{sin x} with respect to x.

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

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

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

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

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

Question 8. Differentiate y = e^{x log x} with respect to x.

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

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

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

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

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

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

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

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

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

Question 18. Differentiate the following with respect to x.

(i) y = x^x √x

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

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

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

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

(vi) y = e^{sin x} + (tan x)^x

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

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

Question 19. Find dy/dx when y = e^x + 10^x + x^x.

Question 20. Find dy/dx when y = xⁿ + n^x+ x^x + nⁿ.

CBSE Class 12 Computer Science

GATE CS & IT 2024

Complete Machine Learning & Data Science Program

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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.5 | Set 1

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

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

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

Question 4. Differentiate with respect to x.

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

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

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

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

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

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

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

Question 12. Differentiate with respect to x.

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

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

Question 15. Differentiate with respect to x.

Question 16. Differentiate with respect to x.

Question 17. Differentiate with respect to x.

Question 18. Differentiate the following with respect to x.

(i) y = xx √x

(ii)

(iii)

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

(v)

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

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

(viii) , for x > 3

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

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

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CBSE Class 12 Computer Science

GATE CS & IT 2024

Complete Machine Learning & Data Science Program

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

Question 2. Differentiate y = x^{sin x} with respect to x.

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

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

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

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

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

Question 8. Differentiate y = e^{x log x} with respect to x.

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

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

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

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

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

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

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

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

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

(i) y = x^x √x

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

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

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

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

(vi) y = e^{sin x} + (tan x)^x

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

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

Question 19. Find dy/dx when y = e^x + 10^x + x^x.

Question 20. Find dy/dx when y = xⁿ + n^x+ x^x + nⁿ.