KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.5 (Q16-Q18)
[Find the derivative of $\frac{x}{\sin x}$ w.r.t sin x]
Sol :
माना $u=\frac{x}{\sin x}$ , v=sinx
Differentiating w.r.t x
$\frac{d u}{d x}=\frac{1 \sin x-x \cdot \cos x}{\sin ^{2} x}$
$\frac{d u}{d x}=\frac{\sin x-x \cos x}{\sin ^{2} x}$..(i)
v=sinx
Differentiating w.r.t x
$\frac{d v}{d x}=\cos x$..(i)
समीकरण (i) मे (ii) से भाग देने पर ,
$\frac{\frac{d y}{d x}}{\frac{dv}{dx}}=\frac{\frac{\sin x-x \cos x}{\sin ^{2} x}}{\cos x}$
$\frac{du}{d v}=\frac{\sin x-x \cos x}{\sin ^{2} x \cos x}$
माना $u=\frac{x}{\sin x}$ , v=sinx
Differentiating w.r.t x
$\frac{d u}{d x}=\frac{1 \sin x-x \cdot \cos x}{\sin ^{2} x}$
$\frac{d u}{d x}=\frac{\sin x-x \cos x}{\sin ^{2} x}$..(i)
v=sinx
Differentiating w.r.t x
$\frac{d v}{d x}=\cos x$..(i)
समीकरण (i) मे (ii) से भाग देने पर ,
$\frac{\frac{d y}{d x}}{\frac{dv}{dx}}=\frac{\frac{\sin x-x \cos x}{\sin ^{2} x}}{\cos x}$
$\frac{du}{d v}=\frac{\sin x-x \cos x}{\sin ^{2} x \cos x}$
Question 17
$\sqrt{\sin \left(1+x^{2}\right)^{2}}$ का 1+x2 के सापेक्ष अवकलन गुणांक ज्ञात करे ।[Find the d.c. of $\sqrt{\sin \left(1+x^{2}\right)^{2}}$ w.r.t $1+x^2$]
Sol :
माना $u=\sqrt{\sin \left(1+x^{2}\right)^{2}}$ ,v=1+x2
Differentiating w.r.t x
$\frac{d u}{d x}=\frac{1}{2 \sqrt{\sin \left(1+x^{2}\right)^{2}}} \times \cos \left(1+x^{2}\right)^{2} \times 2\left(1+x^{2}\right) \cdot 2 x$
$\frac{d u}{d x}=\frac{2 x\left(1+x^{2}\right) \cdot \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}$..(i)
अब,
v=1+x2
Differentiating w.r.t x
$\frac{d v}{d x}=2 x$..(ii)
समीकरण (i) मे (ii) से भाग देने पर ,
$\frac{\frac{d u}{d v}}{\frac{dv}{dx}}=\frac{\frac{2x\left(1+x^{2}\right) \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}}{2x}$
$\frac{d u}{d v}=\frac{\left(1+t^{2}\right) \cdot \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}$
माना $u=\sqrt{\sin \left(1+x^{2}\right)^{2}}$ ,v=1+x2
Differentiating w.r.t x
$\frac{d u}{d x}=\frac{1}{2 \sqrt{\sin \left(1+x^{2}\right)^{2}}} \times \cos \left(1+x^{2}\right)^{2} \times 2\left(1+x^{2}\right) \cdot 2 x$
$\frac{d u}{d x}=\frac{2 x\left(1+x^{2}\right) \cdot \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}$..(i)
अब,
v=1+x2
Differentiating w.r.t x
$\frac{d v}{d x}=2 x$..(ii)
समीकरण (i) मे (ii) से भाग देने पर ,
$\frac{\frac{d u}{d v}}{\frac{dv}{dx}}=\frac{\frac{2x\left(1+x^{2}\right) \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}}{2x}$
$\frac{d u}{d v}=\frac{\left(1+t^{2}\right) \cdot \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}$
Question 18
$\tan ^{-1} \frac{2 x}{1-x^{2}}$ का $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ के सापेक्ष अवकलन गुणांक निकालें जहाँ |x|<1[Find the d.c. of $\tan ^{-1} \frac{2 x}{1-x^{2}}$ w.r.t $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ where |x|<1]
Sol :
माना $u=\tan ^{-1} \frac{2 x}{1-x^{2}}$ $v=\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$
u=2tan-1x,v=2tan-1x
Differentiating w.r.t x
$\frac{d u}{d x}=\frac{2}{1+x^{2}}$..(i)
$\frac{d v}{dx}=\frac{2}{1+t^{2}}$..(ii)
समीकरण (i) मे (ii) से भाग देने पर ,
$\frac{\frac{d u}{d x}}{\frac{d v}{dx}}=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}$
$\frac{d u}{d v}=1$
माना $u=\tan ^{-1} \frac{2 x}{1-x^{2}}$ $v=\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$
u=2tan-1x,v=2tan-1x
Differentiating w.r.t x
$\frac{d u}{d x}=\frac{2}{1+x^{2}}$..(i)
$\frac{d v}{dx}=\frac{2}{1+t^{2}}$..(ii)
समीकरण (i) मे (ii) से भाग देने पर ,
$\frac{\frac{d u}{d x}}{\frac{d v}{dx}}=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}$
$\frac{d u}{d v}=1$
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