KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.4 (Q13-Q15)

Exercise 11.4

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Question 13

यदि (If) (x-y)^m+n=x^m.yⁿ  दिखाएँ कि (show that) $\frac{d y}{d x}=\frac{y}{x}$
Sol :
$(x-y)^{m+n}=x^{m} y^{n}$

Taking log both sides

$\log (x-y)^{m+n}=\log x^{m} \cdot y^{n}$

$(m+n) \cdot \log (x-y)=\log x^{m}+\log y^{n}$

$(m+n) \log (x-y)=m \cdot \log x+n \cdot \log y$

Differentiating w.r.t.x

$(m+n) \cdot \frac{1}{x-y} \times\left[1-\frac{dy}{dx}\right]=m \times \frac{1}{x}+n \times \frac{1}{y} \times \frac{dy}{d x}$

$\frac{m+n}{x-y}-\frac{m+n}{x-y} \frac{dy}{dx}=\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}$

$\frac{m+n}{x-y}-\frac{m}{x}=\left(\frac{m+n}{x-y}+\frac{n}{y}\right) \frac{dy}{d x}$

$\frac{m x+n x-m x+m y}{(x-y) x}=\left(\frac{m y+n y+n x-ny}{(x-y) \cdot y}\right) \frac{dy}{dx}$

$\frac{n x+n y}{x}=\left(\frac{n x+m y}{y}\right) \frac{d y}{d x}$

Question 14

यदि (If) $x=e^{\frac{x}{y}}$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\frac{x-y}{x \log x}$
Sol :
$x=e^{\frac{x}{y}}$

Taking log both sides

$\log x=\log e^{\frac{x}{y}}$

$\log x=\frac{x}{y} \cdot \log e$

$\log x=\frac{x}{y}$

y.log x= x

Differentiating w.r.t.x

$\frac{d y}{d x} \cdot \log x+y \cdot \frac{1}{x}=1$

$\log {x} \frac{d y}{d x}=1-\frac{y}{x}$

$\log x \frac{dy}{dx}=\frac{x-y}{x}$

$\frac{d y}{dx}=\frac{x-y}{x \log x}$


(ii) यदि (If) $x y=e^{x-y}$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\frac{y}{x}\left(\frac{x-1}{y+1}\right)$
Sol :
$x y=e^{x \cdot y}$

Taking log both sides

$\log x \cdot y=\log e^{x-y}$

Differentiating w.r.t.x

$\frac{1}{x}+\frac{1}{y} \times \frac{d y}{d x}=1-\frac{dy}{d x}$

$\frac{1}{y} \frac{d y}{d x}+\frac{d y}{dx}=1-\frac{1}{x}$

$\left(\frac{1}{y}+1\right) \frac{d y}{d x}=1-\frac{1}{x}$

$\left(\frac{1+y}{y}\right) \frac{d y}{dx}=\frac{x-1}{x}$

$\frac{d y}{dx}=\frac{y(x-1)}{x(y+1)}$

Question 15

यदि (If) $y=(\cos x)^{(\cos x)^{(\cos x)}\dots \text{to }\infty }$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\frac{-y^{2} \tan x}{1-y \log \cos x}$
Sol :
$y=(\cos x)^{(\cos x)^{(\cos x)}\dots \text{to }\infty }$

$y=(\cos x)^{y}$

Taking log both sides

$\log y=\log (\cos x)^{y}$

log y=y.log(cos x)

Differentiating w.r.t.x

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d y}{dx} \cdot \log (\cos x)+y \cdot \frac{1}{\cos x} x(-\sin x)$

$\left[\frac{1}{y}-\log (\cos x)\right] \frac{d y}{d x}=-y \tan x$

$\left[\frac{1-y \log(\cos x)}{y}\right] \frac{d y}{d x}=-y \tan x$

$\frac{d y}{dx}=\frac{-y^{2} \tan x}{1-y \log (\cos x)}$