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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.5 | Set 2

  • Last Updated : 26 May, 2021

Question 21. Find dy/dx when y=\frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}     .

Solution:

We have,

=> y=\frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}

=> y=\frac{(x^2-1)^3(2x-1)}{(x-3)^{\frac{1}{2}}(4x-1)^{\frac{1}{2}}}

On taking log of both the sides, we get,



=> log y = log \frac{(x^2-1)^3(2x-1)}{(x-3)^{\frac{1}{2}}(4x-1)^{\frac{1}{2}}}

=> log y = log(x^2-1)^{3}+log(2x-1)-log(x-3)^{\frac{1}{2}}-log(4x-1)^{\frac{1}{2}}

=> log y = 3log(x^2-1)+log(2x-1)-\frac{1}{2}log(x-3)-\frac{1}{2}log(4x-1)

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

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[3log(x^2-1)+log(2x-1)-\frac{1}{2}log(x-3)-\frac{1}{2}log(4x-1)]

=> \frac{1}{y}\frac{dy}{dx}=3(\frac{1}{x^2-1})(2x)+2(\frac{1}{2x-1})-\frac{1}{2}(\frac{1}{x-3})-\frac{1}{2}(\frac{1}{4x-1})(4)

=> \frac{1}{y}\frac{dy}{dx}=\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2(x-3)}-\frac{2}{4x-1}

=> \frac{dy}{dx}=y\left[\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2(x-3)}-\frac{2}{4x-1}\right]



=> \frac{dy}{dx}=\frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}\left[\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2(x-3)}-\frac{2}{4x-1}\right]

Question 22. Find dy/dx when y=\frac{e^{ax}secxlogx}{\sqrt{1-2x}}       .

Solution:

We have,

=> y=\frac{e^{ax}secxlogx}{\sqrt{1-2x}}

=> y=\frac{e^{ax}secxlogx}{(1-2x)^{\frac{1}{2}}}

On taking log of both the sides, we get,

=> log y = log \frac{e^{ax}secxlogx}{(1-2x)^{\frac{1}{2}}}

=> log y = loge^{ax}+logsecx+log(logx)-\frac{1}{2}log(1-2x)

=> log y = ax+logsecx+log(logx)-\frac{1}{2}log(1-2x)

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



=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[ax+logsecx+log(logx)-\frac{1}{2}log(1-2x)]

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

=> \frac{1}{y}\frac{dy}{dx}=a+tanx+\frac{1}{xlogx}+\frac{1}{1-2x}

=> \frac{dy}{dx}=y\left[a+tanx+\frac{1}{xlogx}+\frac{1}{1-2x}\right]

=> \frac{dy}{dx}=\frac{e^{ax}secxlogx}{\sqrt{1-2x}}\left[a+tanx+\frac{1}{xlogx}+\frac{1}{1-2x}\right]

Question 23. Find dy/dx when y = e3x sin 4x 2x.

Solution:

We have 

=> y = e3x sin 4x 2x.

On taking log of both the sides, we get,

=> log y = log (e3x sin 4x 2x)



=> log y = log e3x + log (sin 4x) + log 2x

=> log y = 3x log e + log (sin 4x) + x log 2

=> log y = 3x + log (sin 4x) + x log 2

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

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[3x + log (sin 4x) + x log 2]

=> \frac{1}{y}\frac{dy}{dx}=3+(\frac{1}{sin 4x})(4cos4x) + log2

=> \frac{1}{y}\frac{dy}{dx}=3+4cotx+log2

=> \frac{dy}{dx}=y(3+4cotx+log2)

=> \frac{dy}{dx}=2^xe^{3x}sin4x(3+4cotx+log2)

Question 24. Find dy/dx when y = sin x sin 2x sin 3x sin 4x.

Solution:



We have, 

=> y = sin x sin 2x sin 3x sin 4x

On taking log of both the sides, we get,

=> log y = log (sin x sin 2x sin 3x sin 4x)

=> log y = log sin x + log sin 2x + log sin 3x + log sin 4x

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

=> \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(log sin x + log sin 2x + log sin 3x + log sin 4x)

=> \frac{1}{y}\frac{dy}{dx}=\frac{1}{sinx}(cosx)+\frac{1}{sin2x}(2cos2x)+\frac{1}{sin3x}(3cos3x)+\frac{1}{sin4x}(4cos4x)

=> \frac{1}{y}\frac{dy}{dx}= cotx + 2cot2x + 3cot3x + 4cot4x

=> \frac{dy}{dx}= y(cotx + 2cot2x + 3cot3x + 4cot4x)



=> \frac{dy}{dx}= (sinxsin2x sin3xsin4x)(cotx + 2cot2x + 3cot3x + 4cot4x)

Question 25. Find dy/dx when y = xsin x + (sin x)x.

Solution:

We have, 

=> y = xsin x + (sin x)x

Let u = xsin x and v = (sin x)x. Therefore, y = u + v.

Now, u = xsin x

On taking log of both the sides, we get,

=> log u = log xsin x

=> log u = sin x log x

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

=> \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(sin x log x)

=> \frac{1}{u}\frac{du}{dx}=sin x(\frac{1}{x})+logxcosx

=> \frac{1}{u}\frac{du}{dx}=\frac{sinx}{x}+logxcosx

=> \frac{du}{dx}=u\left[\frac{sinx}{x}+logxcosx\right]

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

Also, v = (sin x)x

On taking log of both the sides, we get,

=> log v = log (sin x)x

=> log v = x log sin x

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



=> \frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(x log sin x)

=> \frac{1}{v}\frac{dv}{dx}=x(\frac{1}{sinx})(cosx)+log(sinx)

=> \frac{1}{v}\frac{dv}{dx}=xcotx+log(sinx)

=> \frac{dv}{dx}=v[xcotx+log(sinx)]

=> \frac{dv}{dx}=(sinx)^x[xcotx+log(sinx)]

Now we have, y = u + v.

=> \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}

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

Question 26. Find dy/dx when y = (sin x)cos x + (cos x)sin x.

Solution:

We have, 



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

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

=> y=e^{cosxlog(sin x)} + e^{sinxlog(cos x)}

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

=> \frac{dy}{dx}=\frac{d}{dx}(e^{cosxlog(sin x)} + e^{sinxlog(cos x)})

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

=> \frac{dy}{dx} = (sinx)cosx[cosxcotx – sinxlog(sinx)] + (cosx)sinx[-tanxsinx + cosxlog(cosx)]

=> \frac{dy}{dx} = (sinx)cosx[cosxcotx – sinxlog(sinx)] + (cosx)sinx[cosxlog(cosx) – tanxsinx]

Question 27. Find dy/dx when y = (tan x)cot x + (cot x)tan x.

Solution:

We have, 



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

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

=> y=e^{cotxlog(tanx)} + e^{tanxlog(cot x)}

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

=> \frac{dy}{dx}=\frac{d}{dx}(e^{cotxlog(tanx)} + e^{tanxlog(cot x)})

=> \frac{dy}{dx}=(e^{cotxlog(tanx)})[cotx(\frac{1}{tanx})(sec^2x)+log(tanx)(-cosec^2x)] + (e^{tanxlog(cot x)})[tanx(\frac{1}{cotx})(-cosec^2x)+log(cotx)(sec^2x)]

=> \frac{dy}{dx}=(tanx)^{cotx}[cot^2x(sec^2x)-log(tanx)(cosec^2x)] + (cotx)^{tanx}[tan^2x(-cosec^2x)+log(cotx)(sec^2x)]

=> \frac{dy}{dx} = (tanx)cotx[cosec2x – log(tanx)(cosec2x)] + (cotx)tanx[-sec2x + log(cotx)(sec2x)]

=> \frac{dy}{dx} = (tanx)cotx[cosec2x – cosec2xlog(tanx)] + (cotx)tanx[sec2xlog(cotx) – sec2x]

Question 28. Find dy/dx when y = (sin x)x + sin−1 √x.

Solution:



We have, 

=> y = (sin x)x + sin−1 √x

=> y=e^{log(sinx)^{x}} + sin^{-1}\sqrt{x}

=> y=e^{xlog(sinx)} + sin^{-1}\sqrt{x}

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

=> \frac{dy}{dx}=\frac{d}{dx}[e^{xlog(sinx)} + sin^{-1}\sqrt{x}]

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

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

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

Question 29. Find dy/dx when 

(i) y = xcos x + (sin x)tan x

Solution:

We have, 

=> y = xcos x + (sin x)tan x

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

=> y=e^{cosxlog(x)} + e^{tanxlog(sinx)}

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

=> \frac{dy}{dx}=\frac{d}{dx}[e^{cosxlog(x)} + e^{tanxlog(sinx)}]

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

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

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

(ii) y = xx + (sin x)x

Solution:



We have, 

=> y = xx + (sin x)x

=> y=e^{logx^{x}} + e^{log(sinx)^{x}}

=> y=e^{xlogx} + e^{xlog(sinx)}

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

=> \frac{dy}{dx}=\frac{d}{dx}[e^{xlogx} + e^{xlog(sinx)}]

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

=> \frac{dy}{dx}=(e^{xlogx})[1+logx] + (e^{xlog(sinx)})[xcotx+log(sinx)]

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

Question 30. Find dy/dx when y = (tan x)log x + cos2 (π/4).

Solution:

We have, 

=> y = (tan x)log x + cos2 (π/4)

=> y=e^{log(tanx)^{logx}} +cos^2(\frac{π}{4})

=> y=e^{logxlog(tanx)} +cos^2(\frac{π}{4})

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

=> \frac{dy}{dx}=\frac{d}{dx}[e^{logxlog(tanx)} +cos^2(\frac{π}{4})]

=> \frac{dy}{dx}=(e^{logxlog(tanx)})[logx(\frac{1}{tanx})(sec^2x)+log(tanx)(\frac{1}{x})] +0

=> \frac{dy}{dx}=(e^{logxlog(tanx)})[\frac{logxsec^2x}{tanx}+\frac{log(tanx)}{x}]

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

Question 31. Find dy/dx when y=x^x+x^{\frac{1}{x}}      .

Solution:



We have,

=> y=x^x+x^{\frac{1}{x}}

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

=> y=e^{xlogx} + e^{\frac{1}{x}logx}

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

=> \frac{dy}{dx}=\frac{d}{dx}[e^{xlogx} + e^{\frac{1}{x}logx}]

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

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

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

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

Question 32. Find dy/dx when y = (log x)x+ xlogx.

Solution:

We have, 

=> y = (log x)x+ xlogx

Let u = (log x)x and v = xlogx. Therefore, y = u + v.

Now, u = (log x)x

On taking log of both the sides, we get,

=> log u = log (log x)x

=> log u = x log (log x)

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

=> \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}[x log (log x)]



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

=> \frac{1}{u}\frac{du}{dx}=\frac{1}{logx}+log(logx)

=> \frac{du}{dx}=u[\frac{1}{logx}+log(logx)]

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

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

=> \frac{du}{dx}=(logx)^{x-1}[1+logxlog(logx)]

Also, v = xlogx

On taking log of both the sides, we get,

=> log v = log xlogx

=> log v = log x (log x)

=> log v = (log x)2

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

=> \frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}[(log x)^2]

=> \frac{1}{v}\frac{dv}{dx}=2(logx)(\frac{1}{x})

=> \frac{dv}{dx}=v(\frac{2logx}{x})

=> \frac{dv}{dx}=x^{logx}(\frac{2logx}{x})

=> \frac{dv}{dx}=2logx.x^{logx-1}

Now, y = u + v

=> \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}

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



Question 33. If x13y7 = (x+y)20, prove that \frac{dy}{dx}=\frac{y}{x}      .

Solution:

We have, 

=> x13y7 = (x+y)20

On taking log of both the sides, we get,

=> log x13y7 = log (x+y)20

=> log x13 + log y7 = log (x+y)20

=> 13 log x + 7 log y = 20 log (x+y)

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

=> 13(\frac{1}{x})+7(\frac{1}{y})(\frac{dy}{dx})=20(\frac{1}{x+y})(1+\frac{dy}{dx})

=> \frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}(1+\frac{dy}{dx})

=> \frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}+\frac{20}{x+y}\frac{dy}{dx}

=> \frac{7}{y}\frac{dy}{dx}-\frac{20}{x+y}\frac{dy}{dx}=\frac{20}{x+y}-\frac{13}{x}

=> \frac{dy}{dx}(\frac{7}{y}-\frac{20}{x+y})=\frac{20x-13x-13y}{x(x+y)}

=> \frac{dy}{dx}(\frac{7x+7y-20y}{y(x+y)})=\frac{7x-13y}{x(x+y)}

=> \frac{dy}{dx}(\frac{7x-13y}{y(x+y)})=\frac{7x-13y}{x(x+y)}

=> \frac{dy}{dx}=\frac{y}{x}

Hence proved.

Question 34. If x16y9 = (x2 + y)17, prove that x\frac{dy}{dx}=2y      .

Solution:

We have, 

=> x16y9 = (x2 + y)17



On taking log of both the sides, we get,

=> log x16y9 = log (x2 + y)17

=> log x16 + log y9 = log (x2 +y)17

=> 16 log x + 9 log y = 17 log (x2 + y)

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

=> 16(\frac{1}{x})+9(\frac{1}{y})(\frac{dy}{dx})=17(\frac{1}{x^2+y})(2x+\frac{dy}{dx})

=> \frac{16}{x}+\frac{9}{y}\frac{dy}{dx}=\frac{34x}{x^2+y}+\frac{17}{x^2+y}\frac{dy}{dx}

=> \frac{9}{y}\frac{dy}{dx}-\frac{17}{x^2+y}\frac{dy}{dx}=\frac{34x}{x^2+y}-\frac{16}{x}

=> (\frac{9}{y}-\frac{17}{x^2+y})\frac{dy}{dx}=\frac{34x^2-16x^2-16y}{x(x^2+y)}

=> (\frac{9x^2+9y-17y}{y(x^2+y)})\frac{dy}{dx}=\frac{34x^2-16x^2-16y}{x(x^2+y)}

=> (\frac{9x^2-8y}{y(x^2+y)})\frac{dy}{dx}=\frac{18x^2-16y}{x(x^2+y)}

=> (\frac{9x^2-8y}{y})\frac{dy}{dx}=\frac{2(9x^2-8y)}{x}

=> \frac{dy}{dx}=\frac{2y}{x}

=> x\frac{dy}{dx}=2y

Hence proved.

Question 35. If y = sin xx, prove that \frac{dy}{dx}=cos(x^x)×x^x(1+logx)     .

Solution:

We have, 

=> y = sin xx 

Let u = xx. Now y = sin u.

On taking log of both the sides, we get,



=> log u = log xx

=> log u = x log x

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

=> \frac{1}{u}\frac{du}{dx}=x(\frac{1}{x})+logx

=> \frac{1}{u}\frac{du}{dx}=1+logx

=> \frac{du}{dx}=u(1+logx)

=> \frac{du}{dx}=x^x(1+logx)

Now, y = sin u

=> \frac{dy}{dx}=cosu\frac{du}{dx}

=> \frac{dy}{dx}=cosu×x^x(1+logx)

=> \frac{dy}{dx}=cosx^x×x^x(1+logx)

Hence proved.

Question 36. If xx + yx = 1, prove that \frac{dy}{dx}=\frac{x^x(1+logx)+y^xlogy}{xy^{x-1}}     .

Solution:

We have, 

=> xx + yx = 1

=> e^{logx^x} + e^{logy^x} = 1

=> e^{xlogx} + e^{xlogy} = 1

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

=> (e^{xlogx})[x(\frac{1}{x})+logx] + (e^{xlogy})[x(\frac{1}{y})(\frac{dy}{dx})+logy] = 0

=> (e^{xlogx})(1+logx) + (e^{xlogy})[(\frac{x}{y})(\frac{dy}{dx})+logy] = 0

=> x^x(1+logx) + y^x[(\frac{x}{y})(\frac{dy}{dx})+logy] = 0

=> x^x(1+logx) + xy^{x-1}\frac{dy}{dx}+y^xlogy = 0

=> xy^{x-1}\frac{dy}{dx} = -[x^x(1+logx)+y^xlogy ]

=> \frac{dy}{dx} = \frac{-[x^x(1+logx)+y^xlogy]}{xy^{x-1}}

Hence proved.

Question 37. If xy × yx = 1, prove that \frac{dy}{dx}=\frac{-y(y+xlogy)}{x(ylogx+x)}     .

Solution:

We have, 

=> xy × yx = 1

On taking log of both the sides, we get,

=> log (xy × yx) = log 1



=> log xy + log yx = log 1

=> y log x + x log y = log 1

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

=> y(\frac{1}{x})+logx(\frac{dy}{dx})+x(\frac{1}{y})(\frac{dy}{dx})+log y = 0

=> \frac{y}{x}+logx(\frac{dy}{dx})+(\frac{x}{y})(\frac{dy}{dx})+log y = 0

=> (\frac{y}{x}+logy)+(logx+\frac{x}{y})(\frac{dy}{dx}) = 0

=> (\frac{y+xlogy}{x})+(\frac{ylogx+x}{y})(\frac{dy}{dx}) = 0

=> (\frac{ylogx+x}{y})(\frac{dy}{dx})= -(\frac{y+xlogy}{x})

=> \frac{dy}{dx}=\frac{-y(y+xlogy)}{x(ylogx+x)}

Hence proved.

Question 38. If xy + yx = (x+y)x+y, find dy/dx.

Solution:

We have, 

=> xy + yx = (x+y)x+y

=> e^{logx^y} + e^{logy^x} = e^{log(x+y)^{x+y}}

=> e^{ylogx} + e^{xlogy} = e^{(x+y)log(x+y)}

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

=> (e^{ylogx})[y(\frac{1}{x})+logx(\frac{dy}{dx})] + (e^{xlogy})[x(\frac{1}{y})(\frac{dy}{dx})+logy]=(e^{(x+y)log(x+y)})[(x+y)(\frac{1}{x+y})(1+\frac{dy}{dx})+log(x+y)(1+\frac{dy}{dx})]

=> e^{ylogx}[\frac{y}{x}+logx(\frac{dy}{dx})] + (e^{xlogy})[(\frac{x}{y})(\frac{dy}{dx})+logy]=(e^{(x+y)log(x+y)})[1+\frac{dy}{dx}+log(x+y)(1+\frac{dy}{dx})]

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

=> \frac{dy}{dx}[x^ylogx+xy^{x-1}-(x+y)^{x+y}(1+log(x+y))]=(x+y)^{x+y}(1+log(x+y))-yx^{y-1}-y^xlogy



=> \frac{dy}{dx}=\frac{(x+y)^{x+y}(1+log(x+y))-yx^{y-1}-y^xlogy}{x^ylogx+xy^{x-1}-(x+y)^{x+y}(1+log(x+y))}

Question 39. If xm yn = 1, prove that \frac{dy}{dx}=-\frac{my}{nx}     .

Solution:

We have, 

=> xm yn = 1

On taking log of both the sides, we get,

=> log (xm yn)= log 1

=> log xm + log yn = log 1

=> m log x + n log y = log 1

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

=> m(\frac{1}{x})+n(\frac{1}{y})(\frac{dy}{dx}) = 0

=> \frac{m}{x}+\frac{n}{y}(\frac{dy}{dx}) = 0

=> \frac{n}{y}(\frac{dy}{dx}) = -\frac{m}{x}

=> \frac{dy}{dx}= -\frac{my}{xn}

Hence proved.

Question 40. If yx = ey−x, prove that \frac{dy}{dx}=\frac{(1+logy)^2}{logy}     .

Solution:

We have, 

=> yx = ey−x

On taking log of both the sides, we get,

=> log yx = log ey−x

=> x log y = (y − x) log e

=> x log y = y − x

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

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

=> \frac{x}{y}\frac{dy}{dx}+logy=\frac{dy}{dx}−1

=> (\frac{x}{y}-1)\frac{dy}{dx}=−1-logy

=> (\frac{x}{y}-1)\frac{dy}{dx}=−(1+logy)

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

=> (\frac{1-1-logy}{1+logy})\frac{dy}{dx}=−(1+logy)

=> \frac{dy}{dx}=−\frac{(1+logy)^2}{-logy}

=> \frac{dy}{dx}=\frac{(1+logy)^2}{logy}

Hence proved.

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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