In Class 12, the concept of differentiation is fundamental to understanding calculus. Chapter 11 of RD Sharma's textbook focuses on differentiation a crucial mathematical technique used to find the rate at which a function is changing at any given point. Exercise 11.5 | Set 3 further explores the various differentiation problems helping students apply the principles they have learned.
Differentiation
Differentiation is a process in calculus that involves finding the derivative of a function. The derivative represents the rate of the change of the function's value with respect to changes in its input. It is a key tool for analyzing functions and solving problems related to motion, optimization, and various real-world applications.
Question 41. If (sin x)y = (cos y)x, prove that \frac{dy}{dx}=\frac{log cosy-ycotx}{logsinx+xtany} .
Solution:
We have,
=> (sin x)y = (cos y)x
On taking log of both the sides, we get,
=> log (sin x)y = log (cos y)x
=> y log (sin x) = x log (cos y)
On differentiating both sides with respect to x, we get,
=>
y[(\frac{1}{sinx})(cosx)]+log(sin x)(\frac{dy}{dx})=x[(\frac{1}{cosy})(-siny)(\frac{dy}{dx})]+log (cos y) =>
ycotx+log(sin x)(\frac{dy}{dx})=-xtany(\frac{dy}{dx})+log (cos y) =>
[log(sin x)+xtany](\frac{dy}{dx})=log (cos y)-ycotx =>
\frac{dy}{dx}=\frac{log cosy-ycotx}{logsinx+xtany} Hence proved.
Question 42. If (cos x)y = (tan y)x, prove that \frac{dy}{dx}=\frac{log tany+ytanx}{logcosx-xsecycosecy} .
Solution:
We have, (cos x)y = (tan y)x
On taking log of both the sides, we get,
=> log (cos x)y = log (tan y)x
=> y log (cos x) = x log (tan y)
On differentiating both sides with respect to x, we get,
=>
y[(\frac{1}{cosx})(-sinx)]+log (cos x)(\frac{dy}{dx})=x[(\frac{1}{tany})(sec^2y)(\frac{dy}{dx})]+log (tan y) =>
-ytanx+log(cos x)(\frac{dy}{dx})=(\frac{xsec^2y}{tany})(\frac{dy}{dx})+log (tan y) =>
-ytanx-log (tan y)=(\frac{xsec^2y}{tany})(\frac{dy}{dx})-log(cos x)(\frac{dy}{dx}) =>
ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(\frac{xsec^2y}{tany})(\frac{dy}{dx}) =>
ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(\frac{x}{cos^2y}×\frac{cosy}{siny})(\frac{dy}{dx}) =>
ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(\frac{x}{cosy}×\frac{1}{siny})(\frac{dy}{dx}) =>
ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(xsecycosecy)(\frac{dy}{dx}) =>
ytanx+log (tan y)=[log(cos x)-(xsecycosecy)]\frac{dy}{dx} =>
[log(cos x)-(xsecycosecy)]\frac{dy}{dx}=ytanx+log (tan y) =>
\frac{dy}{dx}=\frac{ytanx+log (tan y)}{log(cos x)-xsecycosecy} Hence proved.
Question 43. If ex + ey = ex+y, prove that \frac{dy}{dx}+e^{y-x}=0 .
Solution:
We have,
=> ex + ey = ex+y
On differentiating both sides with respect to x, we get,
=>
e^x+e^y(\frac{dy}{dx})=e^{x+y}(1+\frac{dy}{dx}) =>
e^x+e^y(\frac{dy}{dx})=e^{x+y}+e^{x+y}(\frac{dy}{dx}) =>
e^y(\frac{dy}{dx})-e^{x+y}(\frac{dy}{dx})=e^{x+y}-e^x =>
\frac{dy}{dx}(e^y-e^{x+y})=e^{x+y}-e^x =>
\frac{dy}{dx}=\frac{e^{x+y}-e^x}{e^y-e^{x+y}} =>
\frac{dy}{dx}=\frac{e^x+e^y-e^x}{e^y-e^x-e^y} =>
\frac{dy}{dx}=-e^{y-x} =>
\frac{dy}{dx}+e^{y-x}=0 Hence proved.
Question 44. If ey = yx, prove that \frac{dy}{dx}=\frac{(logy)^2}{logy-1} .
Solution:
We have,
=> ey = yx
On taking log of both the sides, we get,
=> log ey = log yx
=> y log e = x log y
=> y = x log y
On differentiating both sides with respect to x, we get,
=>
\frac{dy}{dx}=x(\frac{1}{y})(\frac{dy}{dx})+logy =>
\frac{dy}{dx}=(\frac{x}{y})(\frac{dy}{dx})+logy =>
\frac{dy}{dx}-(\frac{x}{y})(\frac{dy}{dx})=logy =>
\frac{dy}{dx}(1-\frac{x}{y})=logy =>
\frac{dy}{dx}(\frac{y-x}{y})=logy =>
\frac{dy}{dx}=\frac{ylogy}{y-x} =>
\frac{dy}{dx}=\frac{ylogy}{y-\frac{y}{logy}} =>
\frac{dy}{dx}=\frac{ylogy}{\frac{ylogy-y}{logy}} =>
\frac{dy}{dx}=\frac{y(logy)^2}{ylogy-y} =>
\frac{dy}{dx}=\frac{y(logy)^2}{y(logy-1)} Hence proved.
Question 45. If ex+y − x = 0, prove that \frac{dy}{dx}=\frac{1-x}{x} .
Solution:
We have,
=> ex+y − x = 0
On differentiating both sides with respect to x, we get,
=>
e^{x+y}(1+\frac{dy}{dx}) − 1 = 0 =>
e^{x+y}(1+\frac{dy}{dx})= 1 Now, we know ex+y − x = 0
=> ex+y = x
So, we get,
=>
e^{x+y}(1+\frac{dy}{dx})= 1 =>
x(1+\frac{dy}{dx})= 1 =>
1+\frac{dy}{dx}=\frac{1}{x} =>
\frac{dy}{dx}=\frac{1}{x}-1 =>
\frac{dy}{dx}=\frac{1-x}{x} Hence proved.
Question 46. If y = x sin (a+y), prove that \frac{dy}{dx}=\frac{sin^2(a+y)}{sin(a+y)-ycos(a+y)} .
Solution:
We have,
=> y = x sin (a+y)
On differentiating both sides with respect to x, we get,
=>
\frac{dy}{dx}=x[cos(a+y)(\frac{dy}{dx})]+sin(a+y))(1) =>
\frac{dy}{dx}=xcos(a+y)(\frac{dy}{dx})+sin(a+y)) =>
\frac{dy}{dx}-xcos(a+y)(\frac{dy}{dx})=sin(a+y)) =>
\frac{dy}{dx}(1-xcos(a+y))=sin(a+y)) Now we know, y = x sin (a+y)
=>
x=\frac{y}{sin(a+y)} So, we get,
=>
\frac{dy}{dx}(1-\frac{y}{sin(a+y)}cos(a+y))=sin(a+y)) =>
\frac{dy}{dx}(1-\frac{ycos(a+y)}{sin(a+y)})=sin(a+y)) =>
\frac{dy}{dx}(\frac{sin(a+y)-ycos(a+y)}{sin(a+y)})=sin(a+y)) =>
\frac{dy}{dx}=\frac{sin^2(a+y)}{sin(a+y)-ycos(a+y)} Hence proved.
Question 47. If x sin (a+y) + sin a cos (a+y) = 0, prove that \frac{dy}{dx}=\frac{sin^2(a+y)}{sina} .
Solution:
We have,
=> x sin (a+y) + sin a cos (a+y) = 0
On differentiating both sides with respect to x, we get,
=>
x(cos (a+y))(\frac{dy}{dx})+sin(a+y)(1)+sina [-sin(a+y)(\frac{dy}{dx})] = 0 =>
x(cos (a+y))(\frac{dy}{dx})+sin(a+y)-sinasin(a+y)(\frac{dy}{dx}) = 0 =>
\frac{dy}{dx}(xcos(a+y)-sinasin(a+y))+sin(a+y)= 0 =>
\frac{dy}{dx}= \frac{-sin(a+y)}{xcos(a+y)-sinasin(a+y)} Now we know, x sin (a+y) + sin a cos (a+y) = 0
=>
x=\frac{-sinacos(a+y)}{sin(a+y)} So, we get,
=>
\frac{dy}{dx}= \frac{-sin(a+y)}{(\frac{-sinacos(a+y)}{sin(a+y)})cos(a+y)-sinasin(a+y)} =>
\frac{dy}{dx}= \frac{-sin(a+y)}{(\frac{-sinacos^2(a+y)}{sin(a+y)})-sinasin(a+y)} =>
\frac{dy}{dx}= \frac{-sin(a+y)}{(\frac{-sinacos^2(a+y)-sinasin^2(a+y)}{sin(a+y)})} =>
\frac{dy}{dx}= \frac{-sin(a+y)}{\frac{-sina(cos^2(a+y)+sin^2(a+y))}{sin(a+y)}} =>
\frac{dy}{dx}= \frac{-sin(a+y)}{\frac{-sina(1)}{sin(a+y)}} =>
\frac{dy}{dx}= \frac{sin(a+y)}{\frac{sina}{sin(a+y)}} =>
\frac{dy}{dx}= \frac{sin^2(a+y)}{sina} Hence proved.
Question 48. If (sin x)y = x + y, prove that \frac{dy}{dx}=\frac{1-(x+y)ycotx}{(x+y)logsinx-1} .
Solution:
We have,
=> (sin x)y = x + y
On taking log of both the sides, we get,
=> log (sin x)y = log (x + y)
=> y log sin x = log (x + y)
On differentiating both sides with respect to x, we get,
=>
y(\frac{1}{sinx})(cosx)+log sin x(\frac{dy}{dx})=\frac{1}{x + y}(1+\frac{dy}{dx}) =>
ycotx+log sin x(\frac{dy}{dx})=\frac{1}{x + y}+\frac{1}{x+y}\frac{dy}{dx} =>
logsinx(\frac{dy}{dx})-\frac{1}{x+y}\frac{dy}{dx}=\frac{1}{x + y}-ycotx =>
\frac{dy}{dx}(logsinx-\frac{1}{x+y})=\frac{1}{x + y}-ycotx =>
\frac{dy}{dx}(\frac{(x+y)logsinx-1}{x+y})=\frac{1-y(x+y)cotx}{x + y} =>
\frac{dy}{dx}[(x+y)logsinx-1]=[1-y(x+y)cotx] =>
\frac{dy}{dx}=\frac{1-y(x+y)cotx}{(x+y)logsinx-1} Hence proved.
Question 49. If xy log (x+y) = 1, prove that \frac{dy}{dx}=-\frac{y(x^2y+x+y)}{x(xy^2+x+y)} .
Solution:
We have,
=> xy log (x+y) = 1
On differentiating both sides with respect to x, we get,
=>
xy(\frac{1}{x+y})(1+\frac{dy}{dx})+log (x+y)(x\frac{dy}{dx}+y) = 0 =>
\frac{xy}{x+y}(1+\frac{dy}{dx})+xlog(x+y)\frac{dy}{dx}+ylog(x+y)=0 =>
\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+xlog(x+y)\frac{dy}{dx}+ylog(x+y)=0 Now, we know, xy log (x+y) = 1.
=>
log (x+y) = \frac{1}{xy} So, we get,
=>
\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+xlog(x+y)\frac{dy}{dx}+ylog(x+y)=0 =>
\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+x(\frac{1}{xy})\frac{dy}{dx}+y(\frac{1}{xy})=0 =>
\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+\frac{1}{y}\frac{dy}{dx}+\frac{1}{x}=0 =>
\frac{dy}{dx}(\frac{xy}{x+y}+\frac{1}{y})+\frac{xy}{x+y}+\frac{1}{x}=0 =>
\frac{dy}{dx}(\frac{xy^2+x+y}{y(x+y)})+\frac{x^2y+x+y}{x(x+y)}=0 =>
\frac{dy}{dx}(\frac{xy^2+x+y}{y(x+y)})=-\frac{x^2y+x+y}{x(x+y)} =>
\frac{dy}{dx}(\frac{xy^2+x+y}{y})=-\frac{x^2y+x+y}{x} =>
\frac{dy}{dx}=-\frac{y(x^2y+x+y)}{x(xy^2+x+y)} Hence proved.
Question 50. If y = x sin y, prove that \frac{dy}{dx}=\frac{y}{x(1-xcosy)} .
Solution:
We have,
=> y = x sin y
On differentiating both sides with respect to x, we get,
=>
\frac{dy}{dx}=x(cosy)(\frac{dy}{dx})+sin y(1) =>
\frac{dy}{dx}=xcosy\frac{dy}{dx}+sin y =>
\frac{dy}{dx}(1-xcosy)=sin y =>
\frac{dy}{dx}=\frac{sin y}{1-xcosy} Now, we know y = x sin y
=>
siny = \frac{y}{x} So, we get,
=>
\frac{dy}{dx}=\frac{sin y}{1-xcosy} =>
\frac{dy}{dx}=\frac{\frac{y}{x}}{1-xcosy} =>
\frac{dy}{dx}=\frac{y}{x(1-xcosy)} Hence proved.
Question 51. Find the derivative of the function f(x) given by,
f(x) = (1+x) (1+x2) (1+x4) (1+x8) and hence find f'(1).
Solution:
Here we are given,
=> f(x) = (1+x) (1+x2) (1+x4) (1+x8)
On differentiating both sides with respect to x, we get,
=>
f'(x)=(1+x)(1+x^2)\frac{d}{dx}(1+x^8)+(1+x)(1+x^2)(1+x^8)\frac{d}{dx}(1+x^4)+(1+x)(1+x^4)(1+x^8)\frac{d}{dx}(1+x^2)+(1+x^2)(1+x^4)(1+x^8)\frac{d}{dx}(1+x) =>
f'(x)=(1+x)(1+x^2)(1+x^4)(8x^7)+(1+x)(1+x^2)(1+x^8)(4x^3)+(1+x)(1+x^4)(1+x^8)(2x)+(1+x^2)(1+x^4)(1+x^8)(1) Now, the value of f'(x) at 1 is,
=> f'(1) = (1 + 1) (1 + 1) (1 + 1) (8) + (1 + 1) (1 + 1) (1 + 1) (4) + (1 + 1) (1 + 1) (1 + 1) (2) + (1 + 1) (1 + 1) (1 + 1) (1)
=> f'(1) = (2) (2) (2) (8) + (2) (2) (2) (4) + (2) (2) (2) (2) + (2) (2) (2) (1)
=> f'(1) = 64 + 32 + 16 + 8
=> f'(1) = 120
Therefore, the value of f'(1) is 120.
Question 52. If y=log\frac{x^2+x+1}{x^2-x+1}+\frac{2}{\sqrt{3}}tan^{-1}(\frac{\sqrt{3}x}{1-x^2}) , find \frac{dy}{dx} .
Solution:
We are given,
=>
y=log\frac{x^2+x+1}{x^2-x+1}+\frac{2}{\sqrt{3}}tan^{-1}(\frac{\sqrt{3}x}{1-x^2}) On differentiating both sides with respect to x, we get,
=>
\frac{dy}{dx}=\frac{1}{\frac{x^2+x+1}{x^2-x+1}}\frac{d}{dx}(\frac{x^2+x+1}{x^2-x+1})+\frac{2}{\sqrt{3}}\left[\frac{1}{1+(\frac{\sqrt{3}x}{1-x^2})^2}\right]\frac{d}{dx}(\frac{\sqrt{3}x}{1-x^2}) =>
\frac{dy}{dx}=\frac{x^2-x+1}{x^2+x+1}\left[\frac{(x^2-x+1)(2x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}\right]+\frac{2}{\sqrt{3}}\left[\frac{1}{1+\frac{3x^2}{(1-x^2)^2}}\right]\left[\frac{(1-x^2)(\sqrt{3})-\sqrt{3}x(-2x)}{(1-x^2)^2}\right] =>
\frac{dy}{dx}=\frac{1}{x^2+x+1}\left[\frac{2x^3+x^2-2x^2-x+2x+1-2x^3-2x^2-2x+x^2+x+1}{x^2-x+1}\right]+\frac{2}{\sqrt{3}}\left[\frac{1}{\frac{(1-x^2)^2+3x^2}{(1-x^2)^2}}\right]\left[\frac{\sqrt{3}-\sqrt{3}x^2+2\sqrt{3}x^2}{(1-x^2)^2}\right] =>
\frac{dy}{dx}=\frac{-2x^2+2}{x^4+2x^2+1-x^2}+\frac{2}{\sqrt{3}}\left[\frac{1}{\frac{1+x^4-2x^2+3x^2}{(1-x^2)^2}}\right]\left[\frac{\sqrt{3}+\sqrt{3}x^2}{(1-x^2)^2}\right] =>
\frac{dy}{dx}=\frac{-2x^2+2}{x^4+x^2+1}+\frac{2}{\sqrt{3}}\left[\frac{(1-x^2)^2}{1+x^4-2x^2+3x^2}\right]\left[\frac{\sqrt{3}(1+x^2)}{(1-x^2)^2}\right] =>
\frac{dy}{dx}=\frac{-2x^2+2}{x^4+x^2+1}+\frac{2}{\sqrt{3}}\left[\frac{\sqrt{3}(1+x^2)}{1+x^4+x^2}\right] =>
\frac{dy}{dx}=\frac{-2x^2+2}{x^4+x^2+1}+\frac{2(1+x^2)}{x^4+x^2+1} =>
\frac{dy}{dx}=\frac{-2x^2+2+2(1+x^2)}{x^4+x^2+1} =>
\frac{dy}{dx}=\frac{-2x^2+2+2+2x^2}{x^4+x^2+1} =>
\frac{dy}{dx}=\frac{4}{x^4+x^2+1}
Question 53. If y = (sin x − cos x)sin x−cos x, π/4 < x < 3π/4, find \frac{dy}{dx} .
Solution:
We have,
=> y = (sin x − cos x)sin x−cos x
On taking log of both the sides, we get,
=> log y = log (sin x − cos x)sin x−cos x
=> log y = (sin x − cos x) log (sin x−cos x)
On differentiating both sides with respect to x, we get,
=>
\frac{1}{y}\frac{dy}{dx}=(sinx−cosx)(\frac{1}{sinx-cosx})(cosx+sinx)+log (sin x−cos x)(cosx+sinx) =>
\frac{1}{y}\frac{dy}{dx} = (1)(cosx + sinx) + (cosx + sinx)log (sin x − cos x)=>
\frac{1}{y}\frac{dy}{dx} = cosx + sinx + (cosx + sinx)log (sin x − cos x)=>
\frac{1}{y}\frac{dy}{dx} = (cosx + sinx)(1 + log (sin x − cos x))=>
\frac{dy}{dx} = y(cosx + sinx)(1 + log (sin x − cos x))=>
\frac{dy}{dx} = (sinx - cosx)sinx-cosx(cosx + sinx)(1 + log (sin x − cos x))
Question 54. Find dy/dx of function xy = ex-y.
Solution:
We have,
=> xy = ex-y
On taking log of both the sides, we get,
=> log xy = log ex-y
=> log x + log y = (x − y) log e
=> log x + log y = x − y
On differentiating both sides with respect to x, we get,
=>
\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}= 1 −\frac{dy}{dx} =>
\frac{1}{y}\frac{dy}{dx}+\frac{dy}{dx}= 1-\frac{1}{x} =>
(\frac{1}{y}+1)\frac{dy}{dx}= 1-\frac{1}{x} =>
(\frac{1+y}{y})\frac{dy}{dx}=\frac{x-1}{x} =>
\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}
Question 55. Find dy/dx of function yx + xy + xx = ab.
Solution:
We have,
=> yx + xy + xx = ab
=>
e^{logy^x}+e^{logx^y}+e^{logx^x} = a^b =>
e^{xlogy}+e^{ylogx}+e^{xlogx} = a^b On differentiating both sides with respect to x, we get,
=>
(e^{xlogy})[x(\frac{1}{y})(\frac{dy}{dx})+logy]+(e^{ylogx})[y(\frac{1}{x})+logx(\frac{dy}{dx})]+(e^{xlogx})[x(\frac{1}{x})+logx] = 0 =>
(e^{xlogy})[\frac{x}{y}\frac{dy}{dx}+logy]+(e^{ylogx})[\frac{y}{x}+logx(\frac{dy}{dx})]+(e^{xlogx})[1+logx] = 0 =>
(y^x)[\frac{x}{y}\frac{dy}{dx}+logy]+(y^x)[\frac{y}{x}+logx(\frac{dy}{dx})]+(x^x)[1+logx] = 0 =>
(xy^{x-1}+x^ylogy)\frac{dy}{dx}=-x^x(1+logx)-yx^{y-1}-y^xlogy =>
(xy^{x-1}+x^ylogy)\frac{dy}{dx}=-[x^x(1+logx)+yx^{y-1}+y^xlogy] =>
\frac{dy}{dx}=-\frac{x^x(1+logx)+yx^{y-1}+y^xlogy}{xy^{x-1}+x^ylogy}
Question 56. If (cos x)y = (cos y)x, find dy/dx.
Solution:
We have,
=> (cos x)y = (cos y)x
On taking log of both the sides, we get,
=> log (cos x)y = log (cos y)x
=> y log (cos x) = x log (cos y)
On differentiating both sides with respect to x, we get,
=>
y[(\frac{1}{cosx})(-sinx)]+log (cos x)(\frac{dy}{dx})=x[(\frac{1}{cosy})(-siny)(\frac{dy}{dx})]+log(cos y) =>
-ytanx+log (cos x)(\frac{dy}{dx})=-xtany(\frac{dy}{dx})+log(cos y) =>
log (cos x)(\frac{dy}{dx})+xtany(\frac{dy}{dx})=log(cos y)+ytanx =>
[log (cos x)+xtany]\frac{dy}{dx}=log(cos y)+ytanx =>
\frac{dy}{dx}=\frac{log(cos y)+ytanx}{log (cos x)+xtany}
Question 57. If cos y = x cos (a+y), where cos a ≠ ±1, prove that \frac{dy}{dx}=\frac{cos^2(a+y)}{sina} .
Solution:
We have,
=> cos y = x cos (a+y)
On differentiating both sides with respect to x, we get,
=>
(-siny)(\frac{dy}{dx})=x[-sin (a+y)(\frac{dy}{dx})]+cos(a+y)(1) =>
-siny(\frac{dy}{dx})=-xsin (a+y)(\frac{dy}{dx})+cos(a+y) =>
-siny\frac{dy}{dx}+xsin(a+y)\frac{dy}{dx}=cos(a+y) =>
[xsin(a+y)-siny]\frac{dy}{dx}=cos(a+y) =>
\frac{dy}{dx}=\frac{cos(a+y)}{xsin(a+y)-siny} =>
\frac{dy}{dx}=\frac{cos^2(a+y)}{cos(a+y)[xsin(a+y)-siny]} =>
\frac{dy}{dx}=\frac{cos^2(a+y)}{xcos(a+y)sin(a+y)-cos(a+y)siny} =>
\frac{dy}{dx}=\frac{cos^2(a+y)}{cosysin(a+y)-cos(a+y)siny} =>
\frac{dy}{dx}=\frac{cos^2(a+y)}{sin(a+y-y)} =>
\frac{dy}{dx}=\frac{cos^2(a+y)}{sina} Hence proved.
Question 58. If (x-y)e^{\frac{x}{x-y}}=a , prove that y\frac{dy}{dx}+x=2y .
Solution:
We have,
=>
(x-y)e^{\frac{x}{x-y}}=a On differentiating both sides with respect to x, we get,
=>
(x-y)\left[(e^{\frac{x}{x-y}})[\frac{(x-y)-x(1-\frac{dy}{dx})}{(x-y)^2}]\right]+e^{\frac{x}{x-y}}(1-\frac{dy}{dx})=0 =>
\frac{(x-y)-x(1-\frac{dy}{dx})}{(x-y)}+(1-\frac{dy}{dx})=0 =>
(1-\frac{dy}{dx})(1-\frac{x}{x-y})+1=0 =>
(1-\frac{dy}{dx})(\frac{x-y-x}{x-y})+1=0 =>
(1-\frac{dy}{dx})(\frac{-y}{x-y})+1=0 =>
-y+y\frac{dy}{dx}+x-y=0 =>
y\frac{dy}{dx}+x=2y Hence proved.
Question 59. If x=e^{\frac{x}{y}} , prove that \frac{dy}{dx}=\frac{x-y}{xlogx} .
Solution:
We have,
=>
x=e^{\frac{x}{y}} On taking log of both the sides, we get,
=> log x = log
e^{\frac{x}{y}} =>
logx=\frac{x}{y}loge =>
logx=\frac{x}{y} =>
y=\frac{x}{logx} On differentiating both sides with respect to x, we get,
=>
\frac{dy}{dx}=\frac{logx(1)-x(\frac{1}{x})}{(logx)^2} =>
\frac{dy}{dx}=\frac{logx-1}{(logx)^2} We know,
y=\frac{x}{logx} =>
logx=\frac{x}{y} So, we get,
=>
\frac{dy}{dx}=\frac{\frac{x}{y}-1}{(logx)^2} =>
\frac{dy}{dx}=\frac{\frac{x-y}{y}}{(logx)^2} =>
\frac{dy}{dx}=\frac{x-y}{y(logx)^2} =>
\frac{dy}{dx}=\frac{x-y}{(\frac{x}{logx})(logx)^2} =>
\frac{dy}{dx}=\frac{x-y}{xlogx} Hence proved.
Question 60. If y=x^{tanx}+\sqrt{\frac{x^2+1}{2}} , find dy/dx.
Solution:
We have,
=>
y=x^{tanx}+\sqrt{\frac{x^2+1}{2}} =>
y=e^{logx^{tanx}}+e^{log\sqrt{\frac{x^2+1}{2}}} =>
y=e^{tanxlogx}+e^{\frac{1}{2}log(\frac{x^2+1}{2})} On differentiating both sides with respect to x, we get,
=>
\frac{dy}{dx}=(e^{tanxlogx})[tanx(\frac{1}{x})+logx(sec^2x)]+(e^{\frac{1}{2}log(\frac{x^2+1}{2})})(\frac{1}{2})(\frac{2}{x^2+1})(\frac{1}{2})(2x) =>
\frac{dy}{dx}=(e^{tanxlogx})[\frac{tanx}{x}+sec^2xlogx]+(e^{\frac{1}{2}log(\frac{x^2+1}{2})})(\frac{x}{x^2+1}) =>
\frac{dy}{dx}=x^{tanx}[\frac{tanx}{x}+sec^2xlogx]+\sqrt{\frac{x^2+1}{2}}(\frac{x}{x^2+1}) =>
\frac{dy}{dx}=x^{tanx}[\frac{tanx}{x}+sec^2xlogx]+\frac{x}{\sqrt{2(x^2+1)}}
Question 61. If y=1+\frac{\alpha}{(\frac{1}{x}-\alpha)}+\frac{\frac{\beta}{x}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)}+\frac{\frac{\gamma}{x^2}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)(\frac{1}{x}-\gamma)} , find dy/dx.
Solution:
We are given,
=>
y=1+\frac{\alpha}{(\frac{1}{x}-\alpha)}+\frac{\frac{\beta}{x}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)}+\frac{\frac{\gamma}{x^2}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)(\frac{1}{x}-\gamma)} Now we know,
If
y=1+\frac{ax^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{(x-c)} then,\frac{dy}{dx}=\frac{y}{x}[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}] In the given expression, we have 1/x instead of x.
So, using the above theorem, we get,
=>
\frac{dy}{dx}=\frac{\alpha}{(\frac{1}{x}-\alpha)}+\frac{\beta}{(\frac{1}{x}-\beta)}+\frac{\gamma}{(\frac{1}{x}-\gamma)}
Summary
Exercise 11.5 Set 3 in RD Sharma's Class 12 Mathematics textbook further explores optimization problems using differential calculus. This set introduces more advanced applications, often involving multivariable functions, parametric equations, or implicit functions. Students are challenged to apply their knowledge of derivatives to solve complex real-world problems, requiring a deep understanding of both calculus concepts and problem-solving strategies.
Practice Questions:
1. Find the point on the curve y = x² + 2x + 3 that is closest to the point (-1, 4).
2. A box with a square base and open top must have a volume of 32,000 cm³. Find the dimensions of the box that minimize the amount of material used.
3. Find the area of the largest rectangle that can be inscribed in a semicircle of radius 5 units.
4. A particle moves along the curve x = t² - 4t, y = t³ - 6t². Find the point on the curve where its distance from the origin is minimum.
5. Find the dimensions of a cylinder with maximum volume that can be inscribed in a cone of height 12 cm and base radius 6 cm.
6. A wire of length L is cut into two parts. One part is bent into a circle and the other into a square. How should the wire be cut so that the sum of the areas enclosed is maximum?
7. Find the point on the surface xy + yz + zx = 1 that is nearest to the origin.
8. A rectangular sheet of metal of area 300 cm² is to be made into a box by cutting out squares from each corner and folding up the sides. Find the side of the square to be cut out so that the volume of the box is maximum.
9. Find the equation of the line that passes through the point (1, 2) and cuts off the smallest triangle from the first quadrant.
10. A window is in the shape of a rectangle surmounted by an equilateral triangle. The perimeter of the window is 12 m. Find the dimensions of the window to maximize the area.