Differentiation is a fundamental concept in calculus that deals with finding the rate of change of a function with respect to its variable. It allows us to determine how quickly a quantity is changing at any given point. The process involves calculating the derivative of a function, which represents the slope of the tangent line to the function's graph at a specific point. Differentiation has wide-ranging applications in various fields, including physics, engineering, economics, and optimization problems. It enables us to analyze the behavior of functions, find maximum and minimum values, and solve real-world problems involving rates of change. Understanding differentiation is crucial for advanced mathematical analysis and forms the basis for many higher-level mathematical concepts.
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.
Summary
Exercise 11.5 Set 2 in RD Sharma's Class 12 Mathematics textbook continues the focus on optimization problems using differential calculus. This set extends the application of finding maximum and minimum values to more complex scenarios, often involving geometric shapes, economic concepts, or practical situations. Students are required to model these problems mathematically, apply differentiation techniques to find critical points, and interpret the results in the context of the original problem.