KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.4 (Q18-20)

Exercise 11.4

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Question 18

(i) यदि (If) $2^{x}+x^{y}=1$ निकाले (find) $\frac{d y}{d x}$
Sol :
$2^{x}+x^{y}=1$

$2^{x}+e^{\log x^{y}}=1$

$2^{x}+e^{y \cdot \log x}=1$

Differentiating w.r.t.x

$2^{x} \cdot \log 2+e^{y.\log x} \cdot\left[\frac{d y}{d x} \log x+y \cdot \frac{1}{x}\right]=0$

$2^{x} \log {2}+x^{y}\left[\log {x} \frac{d y}{d x}+\frac{y}{x}\right]=0$

$2^{x} \log 2+x^{y} \log _{x} \frac{d y}{d x}+x^{y} \frac{y}{x}=0$

$x^{y} \cdot \log x {\frac{dy}{dx}}=-\left(2^{x} \log x+x^{y} \frac{y}{x}\right)$

$x^{y} \log x \frac{d y}{d x}=-\left(\frac{x 2^{x} \cdot \log x+x^{y} y}{x}\right)$

$\frac{d y}{d x}=-\left(\frac{x \cdot 2^{x} \log x+y \cdot x^{y}}{x^{y+1} \cdot \log {x}}\right)$


(ii) यदि (If) $x^{y}+y^{x}=1$ निकाले ( find ), $\frac{d y}{d x}$
Sol :
$x^{y}+y^{x}=1$

$e^{\log {x^y}}+e^{\log y^{x}}=1$

$e^{y \log x}+e^{x \log y}=1$

Differentiating w.r.t.x

$e^{y \cdot \log x\left[\frac{d y}{dx} \cdot \log x+y \cdot \frac{1}{x}\right]+e^{x} \log y}\left[1 \cdot \log y+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right]=0$

$x^{y}\left[\log x \frac{d y}{dx}+\frac{y}{x}\right]+y^{x}\left[\log y+\frac{x}{y} \frac{d y}{dx}\right]=0$

$x^{y} \cdot \log x \frac{d y}{dx} \frac{y x^{y}}{x}+y^{x} \log{ y}+\frac{x y^{x}}{y} \cdot \frac{d y}{d x}=0$

$\left[x^{y} \log x+\frac{x y^{x}}{y}\right] \frac{d y}{d x}=-\left[\frac{y x^{y}}{x}+y^{x} \log y\right]$

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

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

Question 19

(i) यदि (If) $y=\frac{x^{2} \sqrt{4 x+3}}{(3 x+1)^{2}}$ निकाले (find) $\frac{d y}{d x}$
Sol :
$y=\frac{x^{2} \sqrt{4 x+3}}{(3 x+1)^{2}}$

Taking log both sides

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

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

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

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

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d{y}}{d{x}}=2 \times \frac{1}{x}+\frac{1}{2} \times \frac{1}{4 x+3} \times 4+2\times \frac{1}{3 x+1} \times 3$

$\frac{d y}{d x}=y^{2}\left[\frac{1}{x}+\frac{1}{4 x+3}-\frac{3}{3 x+1}\right]$

$\frac{d y}{d x}=\frac{x^{2} \sqrt{4 x+3}}{(3 x+1)^{2}} \cdot 2\left[\frac{1}{x}+\frac{1}{4 x+3}-\frac{3}{3 x+1}\right]$


(ii) यदि (If) $y=\frac{2(x-\sin x)^{\frac{3}{2}}}{\sqrt{x}}$ निकाले (find) $\frac{d y}{d x}$
Sol :
$y=\frac{2\left(x-\sin x\right)^{\frac{3}{2}}}{\sqrt{x}}$

Taking log both sides

$\log y=\log \frac{2[x-\sin x]^{\frac{3}{2}}}{x^{1 / 2}}$

$\log y=\log {2}(x-\sin x)^{\frac{3}{2}}-\log x^{\frac{1}{2}}$

$\log y=\log 2+\frac{3}{2} \log (x-\sin x)-\frac{1}{2} \log x$

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{dy}{dx}=0+\frac{3}{2} \times \frac{1}{x-\sin x} \times[1-\cos x]-\frac{1}{2} \times \frac{1}{x}$

$\frac{1}{y} \frac{d y}{d x}=\frac{3\left(1-\cos x\right)}{2(x-\sin x)}-\frac{1}{2 x}$

$\frac{d y}{d x}=y\left[\frac{3 x-3 x \cos x-x+\sin x}{2 x(x-\sin x)}\right]$

$\frac{d y}{d x}=\frac{2(x-\sin x)^{\frac{3}{2}}}{\sqrt{x}}\left[\frac{2 x-3 x \cos x+\sin x}{2 x\left(x-\sin x\right)}\right]$

Question 20

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें ।

[Differentiate the following functions w.rt.x]