KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.3 (Q4-Q6)

Exercise 11.3

---

---

Question 4

$\frac{d y}{d x}$ ज्ञात करें यदि

[Find $\frac{dy}{dx}$ , if]

(i) $x^{m} y^{n}=(x-y)^{m+n}$
Sol :
Differentiating with respect to x

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

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

$\left[\because x^{m} y^{n}=(x-y)^{m+n}\right]$

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

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

$\left[\frac{n}{y}+\frac{m+n}{x-y}\right] \frac{d y}{d x}=\frac{m+n}{x-y}-\frac{m}{x}$

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

$\left[\frac{n x+m y}{y}\right] \frac{d y}{dx}=\frac{n x+m y}{x}$

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


(ii) $x^{3} y^{4}=(x-y)^{7}$
Sol :
Differentiating with respect to x

$3 x^{2} y^{4}+x^{3} \cdot 4 y^{3} \cdot \frac{d y}{dx}=7(x-y)^{7-1} \cdot\left[1-\frac{d y}{d x}\right]$

$x^{3} y^{4}\left[\frac{3}{x}+\frac{4}{y} \frac{d y}{d x}\right]=\frac{7(x-y)^{7}}{x-y}\left[1-\frac{dy}{d x}\right]$

$\left[\because x^{3} y^{4}=(x-y)^{7}\right]$

$\frac{3}{x}+\frac{4}{y} \frac{d}{d}=\frac{7}{x-y}-\frac{7}{x-y} \cdot \frac{dy}{dx}$

$\frac{4}{y} \frac{d y}{d x}+\frac{7}{x-y} \frac{d y}{d x}=\frac{7}{x-y}-\frac{3}{x}$

$\left[\frac{4}{y}+\frac{7}{x-y}\right] \frac{dy}{d x}=\frac{7}{x-y}-\frac{3}{x}$

$\left[\frac{4 x-4 y+7 y}{y(x-y)}\right] \frac{d y}{d x}=\frac{7 x-3 x+3 y}{x(x-y)}$

$\left[\frac{4 x+3 y}{y(x-y)}\right] \frac{d y}{d x}=\frac{4 x+3 y}{x(x-y)}$

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


(iii) $x^{2} y^{2}=(x-y)^{4}$
Sol :
Differentiating with respect to x

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

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

$\left[\because x^{2} y^{2}=(x-y)^{4}\right]$

$\frac{2}{x}+\frac{2}{y} \frac{d y}{d x}=\frac{4}{x-y}-\frac{4}{x-y} \frac{dy}{dx}$

$\frac{2}{y} \frac{d y}{d x}+\frac{4}{x-y} \frac{d y}{dx}=\frac{4}{x-y}-\frac{2}{x}$

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

$\left[\frac{2 x-2 y+4 y}{y(x-y)]}\right] \frac{d y}{d x}=\frac{4 x-2 x+2 y}{x(x-y)}$

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

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


(iv) $x^{2} y=(2 x+3 y)^{3}$
Sol :
Differentiating with respect to x

$2 x \cdot y+x^{2} \cdot \frac{d y}{dx}=3(2 x+3 y)^{3-1} \cdot\left(2+\frac{3 dy}{dx}\right)$

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

$\left[\because x^{2} y=(2 x+3 y)^{3}\right]$

$\frac{2}{x}+\frac{1}{y} \frac{dy}{dx}=\frac{6}{2 x+3 y}+\frac{9}{2 x+3y}- \frac{dy}{dx}$

$\left[\frac{1}{y}-\frac{9}{2 x+3 y}\right] \frac{d y}{d x}=\frac{6}{2 x+3 y}-\frac{2}{x}$

$\left[\frac{2 x+3 y-9 y}{y(2 x+3 y)}\right] \frac{d y}{d x}=\frac{6 x-4 x-6 y}{x(2 x+3y)}$

$\left[\frac{2 x-6 y}{y(2 x+3 y)}\right] \frac{d y}{d x}=\frac{2 x-6 y}{x(2 x+3 y)}$

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


(v) $x^3y^4=(x+y)^7$
Sol :