Exercise 11.4
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]
(i) $y=x^{\frac{1}{x}}$
Sol :
Taking log both sides
$\log y=\log x^{\frac{1}{x}}$
$\log y=\frac{1}{x} \cdot \log x$
Differentiating w.r.t.x
$\frac{1}{y} \times \frac{dy}{dx}=\frac{-1}{x^{2}} \log {x}+\frac{1}{x} \times \frac{1}{x}$
$\frac{d y}{d x}=y\left[\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right]$
$=\frac{x^{\frac{1}{x}}}{x^{2}}\left[1-\log x\right]$
(ii) y=x
y
Sol :
Taking log both sides
log y=log x
y
log y=y.log x
Differentiating w.r.t.x
$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d y}{d x} \cdot \log x+y \cdot \frac{1}{x}$
$\frac{1}{y} \cdot \frac{d y}{d x}-\log x \frac{d y}{d x}=\frac{y}{x}$
$\left(\frac{1}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}$
$\left(\frac{1-y \log x}{y}\right) \frac{d y}{d x}=\frac{y}{x}$
$\frac{d y}{d x}=\frac{y^{2}}{x(1-y \log x)}$
(iii) y=x
log x
Sol :
Taking log both sides
log y=log x
log x
log y=log x.log x
log y=(log x)
2
Differentiating w.r.t.x
$\frac{d y}{d x}=y \cdot 2 \frac{(\log x)}{x}$
$=x^{\log x} \cdot \frac{2\left(\log x\right)}{x}$
(iv) $y=x^{x+\frac{1}{x}}$
Sol :
Taking log both sides
$\log {y}=\log {x}\left(x+\frac{1}{x}\right)$
$\log {y}=\left(x+\frac{1}{x}\right) \cdot \log x$
Differentiating w.r.t.x
$\frac{1}{y}\times \frac{d y}{d x}=\left(1-\frac{1}{x^{2}}\right) \log x+\left(x+\frac{1}{x}\right) \cdot \frac{1}{x}$
$\frac{d y}{dx}=y\left[1+\frac{1}{x^{2}}+\left(1-\frac{1}{x^{2}}\right) \log x\right]$
$=x^{x+\frac{1}{x}}\left[1+\frac{1}{x^{2}}+\left(1-\frac{1}{x^{2}}\right) \log {x}\right]$
(v) $y=x^{\cos ^{-1} x}$
Sol :
$y=x^{\cos ^{-1} x}$
Taking log both sides
$\log {y}=\log x^{\cos^{-1}x}$
$\log y=\cos ^{-1} x \cdot \log x$
Differentiating w.r.t.x
$\frac{1}{y}. \frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \log x+\cos^{-1} x\times\frac{1}{x}$
$\frac{d y}{dx}=y\left[\frac{1}{x} \cos ^{-1} x-\frac{1}{\sqrt{1-x^{2}}} \log x\right]$
$=x \cos^{-1} x\left[\frac{1}{x} \cos ^{-1} x-\frac{1}{\sqrt{1-x^{2}}} \log x\right]$
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