KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.4 (Q7-Q9)

Exercise 11.4

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Question 7

यदि (If) x=ylog (xy) दिखाएँ कि (show that) $\frac{d y}{d x}=\frac{y(x-y)}{x(x+y)}$
Sol :
x=ylog (xy)

x=y[logx+logy]

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

Differentiating w.r.t.x

$1=\frac{dy}{dx}\left[\log x+\log y\right]+y \cdot\left[\frac{1}{x}+\frac{1}{y} \cdot \frac{d y}{d x}\right]$

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

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

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

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

Question 8

$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\text{to }\infty}}}$ सिद्ध करें कि ( prove that ) $\frac{d y}{d x}=\frac{1}{x(2 y-1)}$
Sol :
$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\text{to }\infty}}}$

$y=\sqrt{\log x+y}$

Squaring both sides

$y^{2}=\log x+y$

Differentiating w.r.t.x

$2 y \cdot \frac{d y}{d x}=\frac{1}{x}+\frac{d y}{d x}$

$2 y \frac{d y}{d x}-\frac{dy}{dx}=\frac{1}{x}$

$(2 y-1) \frac{d y}{d x}=\frac{1}{x}$

$\frac{dy}{dx}=\frac{1}{x(2 y-1)}$

Question 9

$\frac{d y}{d x}$ निकालें यदि 

[Find $\frac{d y}{d x}$ if]