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

  • Last Updated : 26 May, 2021

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)




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