Exercise 11.4
Question 16
निम्नलिखित फलनों को x के सापेक्ष अवकलित करें ।
[Differentiate the following functions w.rt. x]
KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.4 (Q16-Q17)
(i) $(\log x)^{x}+x^{\log x}$
Sol :
Let $y=u+v=(\log x)^{x}+x^{\log _{x}}$
$u=(\log x)^{x}$
Taking log both sides
$\log u=\log \left(\log {x}\right)^{x}$
log u=x log(log x)
Differentiating w.r.t.x
$\frac{1}{u} \cdot \frac{d u}{d x}=1.\log (\log x)+x \times \frac{1}{\log x} \times \frac{1}{2}$
$\frac{d u}{d x}=u\left[\log (\log x)+\frac{1}{\log x}\right]$
$\frac{d u}{d x}=(\log x)^{x}\left[\log \left( \log x\right)+\frac{1}{\log x}\right]$
$\frac{du}{dx}=\left(\log x\right)^{x}\left[\frac{\left[\log x \log (\log x)+1\right.}{\log x}\right]$
$\frac{du}{dx}=\left(\log x\right)^{x-1}\left[1+\log {x} \cdot \log (\log x)\right]$
Now ,
$v=x^{\log x}$
Taking log both sides
$\log v=\log {x} ^{\log x}$
log v=log x.log x
$\log v=(\log x)^{2}$
Differentiating w.r.t.x
$\frac{1}{v} \cdot \frac{d v}{d x}=2 \log x \times \frac{1}{x}$
$\frac{d v}{d x}=v \cdot \frac{2 \log x}{x}$
$\frac{d v}{d x}=x^{\log x} \cdot \frac{2 \log x}{x}$
$\frac{d v}{dx}=2 x^{\log x-1} \cdot \log x$
∵ y=u+v
Differentiating w.r.t.x
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 x^{\log x-1} \log x$
(ii) $x^{\sin x}+(\sin x)^{\cos x}$
Sol :
Let y=u+v=$x^{\sin x}+(\sin x)^{\cos x}$
$u=x^{\sin x}$
Taking log both sides
$\log u=\log x ^{\sin x}$
log u=sin x.log x
Differentiating w.r.t.x
$\frac{1}{u}\times \frac{d u}{d x}=\cos x \cdot \log x+\sin x \times \frac{1}{x}$
$\frac{du}{d x}=4\left[\frac{\sin x}{x}+\cos x \cdot \log x\right]$
$\frac{du}{dx}=x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]$
Now , $v=(\sin x)^{\cos x}$
Taking log both sides
$\log v=\log (\sin x)^{\cos x}$
$\log v=\cos x \cdot \log (\sin x)$
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d x}=-\sin x \log (\sin x)+\cos x\times \frac{1}{\sin x} \times \cos x$
$\frac{d{v}}{dx}=v[\cos x \cdot \cot x-\sin x \log (\sin x)]$
$\frac{d v}{d x}=(\sin x)^{\cos x}[\cos x \cot x-\sin x \log (\sin x)]$
∵ y=u+v
Differentiating w.r.t.x
$\frac{d y}{d x}=x^{\sin x}\left[\frac{\sin x}{2}+\cos x \log x\right]+(\sin x)^{\cos x}[\cos x.\cot x-\sin x\log(\sin x) ]$
(iii) $x^{x}-2 \sin x$
Sol :
Let y=u-v=$x^{x}-2 \sin x$
$u=x^{x}$
Taking log both sides
$\log u=\log x^{x}$
log u=x.log x
Differentiating w.r.t.x
$\frac{1}{u} \cdot \frac{d y}{dx}=1 \cdot \log x+x \times \frac{1}{x}$
$\frac{d u}{d x}=u[1+\log x]$
$\frac{d u}{d x}=x^{x}\left[1+\log x\right]$
Now , $v=2^{\sin x}$
Taking log both sides
$\log v=\log 2^{\sin x}$
log v=sin x .log 2
Differentiating w.r.t.x
$\frac{1}{v} \cdot \frac{d v}{d x}=\cos x \cdot \log {2}$
$\frac{d v}{dx}=2^{\sin x} \cdot \cos x \log 2$
∵y=u-v
Differentiating w.r.t.x
$\frac{d y}{dx}=\frac{d u}{dx}-\frac{d v}{d x}$
$\frac{d y}{d x}=x^{2}(1+\log x)-2^{\sin x} \cos x\log 2$
(iv) $(\sin x)^{x}+\sin ^{-1} \sqrt{x}$
Sol :
Let y=$(\sin x)^{x}+\sin ^{-1} \sqrt{x}$
$u=(\sin x)^{x}$
Taking log both sides
$\log u=\log (\sin x)^{x}$
log u=x.log(sin x)
Differentiating w.r.t.x
$\frac{1}{u} \times \frac{d y}{d x}=1 \cdot \log (\sin x)+x \frac{1}{\sin x} \times \cos x$
$\frac{du}{dx}=u[x \cot x+\log (\sin x)]$
$\frac{du}{dx}=(\sin x)^{x}\left[x \cot x+\log \left(\sin x\right)\right]$
Now , $u=\sin ^{-1} \sqrt{2}$
Differentiating w.r.t.x
$\frac{d v}{d u}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \times \frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{x} \sqrt{1-2}}$
$\frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$
∵ y=u+v
Differentiating w.r.t.x
$\frac{d y}{d x}=\frac{d y}{d x}+\frac{d v}{d x}=(\sin x)^{x}[x \cot x+\log (\sin x)]+\frac{1}{2 \sqrt{x-x^{2}}}$
(v) $x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
Sol :
Let y=u+v=$x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
$u=x^{x \cos x}$
Taking both sides log
$\log u=\log x^{x \cos x}$
$\log u=x \cos x \cdot \log {x}$
Differentiating w.r.t.x
$\frac{1}{u} \times \frac{d u}{d x}=1 \cdot \cos x \log x+x(-\sin x) \log x + x \cos x \times \frac{1}{x}$
$\frac{du}{dx}=u\left[\cos x+\cos x \log x-x \sin x \log x\right]$
$\frac{d u}{d x}=x^{x\cos x} [\cos x(1+\log x)-x \sin x \log x]$
Now , $v=\frac{x^{2}+1}{x^{2}-1}$
Differentiating w.r.t.x
$\frac{d v}{d x}=\frac{2 x\left(x^{2}-1\right)-\left(x^{2}+1\right) \cdot 2 x}{\left(x^{2}-1\right)^{2}}$
$=\frac{2 x^{3}-2 x-2 x^{3}-2 x}{\left(x^{2}-1\right)^{2}}$
$=\frac{-4 x}{\left(x^{2}-1\right)^{2}}$
∵y=u+v
Differentiating w.r.t.x
$\frac{d y}{dx}=\frac{d y}{d x}+\frac{d v}{d x}$
$=x^{x \cos x}[\cos x(1+\log x)-x \sin x \log x]$
(vi) $(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$
Sol :
Let y=$(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$
$u=(x \cos x)^{x}$
Taking log both sides
$\log u=\log (x \cos x)^{x}$
$\log u=x \cdot \log (x \cos x)$
Differentiating w.r.t.x
$\frac{1}{u} \times \frac{d y}{d x}=1 . \log \left(x \cos x\right)+x \cdot \frac{1}{x \cos x}[1.\cos x+x(-\sin x)]$
$\frac{du}{dx}=u[\log (x \cos x)+1-x \tan x]$
$\frac{d y}{d x}=(x \cos x)^{x}[1-x \tan x+\log (x \cos x)]$
Now , $v=(x \sin x)^{\frac{1}{x}}$
Taking log both sides
$\log v=\log (x \sin x)^{\frac{1}{x}}$
$\log v=\frac{1}{x} \log (x \sin x)$
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d x}=\frac{\frac{1}{x \sin x}[1 . \sin x+x \cos x] x-\log {y}(x \sin 3 x).1]}{x^2}$
$\frac{1}{v} \times \frac{du}{dx}=\frac{1+x \cot x-\log (x\sin x)}{x^{2}}$
$\frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1+x \cot x-\log (x \sin x)]}{x^{2}}\right]$
y=u+v
Differentiating w.r.t.x
$\frac{d y}{dx}=\frac{d y}{d x}+\frac{d v}{d x}$
$=(x \cos x)^x[1-x \tan x+\log(x \cos x)]+(x \sin x)^{\frac{1}{x}}\left[\frac{1+x\cot x-\log (x\sin x)}{x^{2}}\right]$
(vii) $x^{x^{2}-3}+(x-3)^{x^{2}}, x>3$
Sol :
Let y=u+v=$x^{x^{2}-3}+(x-3)^{x^{2}}$
$u=x^{x^{2}-3}$
Taking log both sides
$\log u=\log x^{x^2}-3$
$\log u=\left(x^{2}-3\right) \cdot \log x$
Differentiating w.r.t.x
$\frac{1}{u} \cdot \frac{d y}{d x}=2 x \cdot \log x+\left(x^{2}-3\right) \times \frac{1}{x}$
$\frac{d y}{dx}=u\left[\frac{x^{2}-3}{x}+2 x \log x\right]$
$\frac{d y}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]$
Now , $v=(x-3)^{x^{2}}$
Taking log both sides
$\log v=\log {(x-3)}^{x^{2}}$
$\log v=x^{2} \cdot \log (x-3)$
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d x}=2 x \cdot \log (x-3)+x^{2} \frac{1}{x-3}$
$\frac{dv}{dx}=v\left[\frac{x^{2}}{x-3}+2 x \log (x-3)\right]$
$\frac{d v}{dx}=(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2 x \log (x-3)\right]$
∵ y=u+v
Differentiating w.r.t.x
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\frac{d y}{dx}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]+(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2x \log(x-3)\right]$
Question 17
Sol :
Let $u=x^{x}, v=x^{\frac{1}{x}}$ and
y=u+v
$u=x^{x}$
Taking log both sides
$\log u=\log x^x$
$\log u=x \cdot \log {x}$
Differentiating w.r.t.x
$\frac{1}{u} \times \frac{d y}{d x}=1 \cdot \log x+x \times \frac{1}{x}$
$\frac{d u}{d x}=x^{x}[\log x+1]$
$v=x^{\frac{1}{x}}$
Taking log both sides
$\log v=\log {x} \frac{1}{x}$
$\log v=\frac{1}{x} \cdot \log x$
$\log v=\frac{\log {x}}{x}$
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d u}=\frac{\frac{1}{x} \times x-\lg x \cdot 1}{x^{2}}$
$\frac{d v}{dx}=x^{\frac{1}{x}}\left[\frac{1-\log x}{x^{2}}\right]$
y=u+v
Differentiating w.r.t.x
$\frac{d y}{dx}=\frac{d u}{dx}+\frac{d v}{dx}$
$=x^{x}(\log x+1)+x^{\frac{1}{x}}\left(\frac{1-\log x}{x^{2}}\right)$
(ii) यदि (If) $y=x+x^{\frac{1}{x}}$ (find) $\frac{dy}{dx}$ निकालें।
Sol :
(iii) यदि (If) $y=(\sin x)^{\cos x}+(\cos x)^{\sin x}$ निकाले (Find) $\frac{d y}{d x}$
Sol :
Let $u=(\sin x)^{\cos x}, \quad v=(\cos x)^{\sin x}$
y=u+v
$u=(\sin x)^{\cos x}$
Taking log both sides
$\log u=\log (\sin x)^{\cos x}$
log u=cos x. log (sin x)
Differentiating w.r.t.x
$\frac{1}{u} \times \frac{du}{dx}=-\sin x \log (\sin x)+\cos x \times \frac{1}{x}\times \cos x$
$\frac{du}{dx}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)]$
$\frac{d u}{dx}=(\sin x)^{\cos x}\left[\cot x-\log (\tan x)^{\sin x}\right]$
Now , $v=(\cos x)^{\sin x}$
Taking log both sides
$\log v=\log (\cos x)^{\sin x}$
$\log v=\sin x \cdot \log (\cos x)$
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d x}=\cos x\log(\sin x)+\sin x \frac{1}{\cos x} \times(-\sin x)$
$\frac{d v}{d x}=(\cos )^{\sin x}\left[\log (\cos x)^{\cos x}-\tan x \sin x\right]$
y=u+v
Differentiating w.r.t.x
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{dx}$
$\frac{d y}{dx}=(\sin x)^{\cos x}\left[\cot x \cos x-\log (\sin x)^{\sin x}\right]+\left(\cos\right)^{\sin x }\left[\log \left(\cos x\right)^{\cos x}-\tan x \sin x\right]$
(iv) यदि (If) $y=x^{\tan x}+(\tan x)^{\cot x}$ निकाले (find) $\frac{d y}{d x}$
Sol :
Let $u=x^{\tan x}$ , $v=(\tan x)^{\cot x}$ , $y=u+v$
$u=x^{\tan x}$
Taking log both sides
$\log u=\log x^{ \tan x}$
log u=tan x.log x
Differentiating w.r.t.x
$\frac{1}{u} \cdot \frac{du}{dx}=\sec ^{2} x \cdot \log {x}+\tan x \times\frac{1}{x}$
$\frac{d y}{d x}=x^{\tan x}\left[\frac{\tan x}{2}+\sec^{2} x \cos x\right]$
Now , $v=(\tan x)^{\operatorname{cot} x}$
Taking log both sides
$\log {v}=\log (\tan x)^{\cot x}$
log v=cot x.log (tan x)
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d x}=-\operatorname{cosec}^{2} x \cdot \log (\tan x)+\cot x \times\frac{1}{\tan x} \times \operatorname{sec}^{2} x$
$\frac{1}{v} \times \frac{d v}{d x}=\frac{\cos x}{\sin x} \times \frac{1}{\frac{\sin x}{\cos x}} \times \frac{1}{\cos ^{2} x}-\operatorname{cosec}^{2} x \log \left(\tan x\right)$
$\frac{d v}{dx}=(\tan x)^{\cot x}\left[\operatorname{cosec}^{2} x-\operatorname{cosec}^{2} x \log (\tan x)\right]$
$\frac{dv}{dx}=(\tan x)^{\cot x} \operatorname{cosec}^{2} x[1-\log (\tan x)]$
y=u+v
Differentiating w.r.t.x
$\frac{d y}{dx}=\frac{d y}{dx}+\frac{d y}{d x}$
$\frac{d y}{dx}=x^{\tan x}\left[\frac{\tan x}{x}+\sec ^{2} x \log x\right]+(\tan x)^{\cot x} \cdot \text{cosec x}^{2} x(1-\log (\tan x)]$
(v) यदि (If) $y=(\tan x)^{\cot x}+(\cot x)^{\tan x}$ निकाले (find) $\frac{d y}{d x}$
Sol :
Let $u=(\tan x)^{\cot x}$ $ v=(\cot x)^{\tan x}$
y=u+v
$u=(\tan x)^{\cot x}$
Taking log both sides
$\log u=\log (\tan x)^{\cot x}$
log u=cot x.log(tan x)
Differentiating w.r.t.x
$\frac{1}{u} \times \frac{d y}{dx}=-\operatorname{cosec}^{2} x \cdot \log (\tan x)+\cot x \times \frac{1}{\tan x} \times \operatorname{sec}^{2} x$
$\frac{d u}{dx}=u\left[\frac{\cos x}{\sin x} \times \frac{1}{\frac{\sin x}{\cos x}} \times \frac{1}{\cos ^{2} x}-\operatorname{cosec}^{2} x \log (\tan x)\right]$
$\frac{d u}{d x}=(\tan x)^{\cot x}.\text{cosec}^{2}x[1-\log(\tan x)]$
Now , $v=(\cot x)^{\tan x}$
Taking log both sides
$\log v=\log \left(\cot {x}\right)^{\tan x}$
$\log v=\tan x \cdot \log (\cot x)$
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{dx}=\sec ^{2} x \log (\operatorname{cot x})+\tan x \cdot \frac{1}{\operatorname{cot} x} \times\left(-\text{cosec} ^{2} x\right)$
$\frac{d v}{d x}=v\left[\sec ^{2} x \log (\cot x)-\frac{\sin x}{\cos x} \times \frac{1}{\frac{\cos x}{\sin x}} \times \frac{1}{\sin^{2}x}\right.$
$\frac{d v}{d x}=(\cot x)^{\tan x} \sec ^{2} x[\log (\cot x)-1]$
y=u+v
Differentiating w.r.t.x
$\frac{d y}{dx}=\frac{d u}{d x}-\frac{d v}{d x}$
$\frac{d y}{d x}=(\tan x)^{\cot x} \operatorname{cosec}^{2} x(1-\log (\tan x)]+(\cot x)^{\tan x}.\sec^2x[\log(\cot x)-1]$
(vi) यदि (If) $y=x^{x}+(\sin x)^{\cot x}$ निकाले (Find) $\frac{d y}{d x}$
Sol :
Let $u=x^{x}$ $v=(\sin x)^{\cos x}$
y=u+v
$u=x^{x}$
Taking log both sides
$\log u=\log x^{x}$
log u=x.logx
Differentiating w.r.t.x
$\frac{1}{u} \cdot \frac{d y}{dx}=1 \cdot \log x+x \cdot \frac{1}{x}$
$\frac{du}{dx}=x^{2}\left[1+\log {x}\right]$
Now , $v=(\sin x)^{\operatorname{cot x}}$
log v=cot x.log(sin x)
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d x}=-\text{cosec} ^{2} x \cdot \log (\sin x)+\cot x \frac{1}{\sin x} \times \cos x$
$\frac{d v}{dx}=v\left[\cot^{2} x-\text{cosec} ^{2} x \log (\sin x)\right]$
$\frac{d v}{dx}=(\sin x)^{\cot x}\left[\cot ^{2} x-\text{cosec} ^{2} x \log (\sin x)\right]$
∵ y=u+v
Differentiating w.r.t.x
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\frac{d y}{dx}=x^{x}\left(1+\log x\right)+(\sin x)^{\cot x}[\cot^2 x-\text{cosec}^2x.\log(\sin x)]$
(vii) यदि(If) $y=(x)^{\cos x}+(\cos x)^{\sin x}$ निकाले (find) $\frac{d y}{d x}$
Sol :
Let $u=x^{\cos x}$ $ v=(\cos x)^{\sin x}$
y=u+v
$u=x^{\cos x}$
Taking log both sides
$\log u=\log x^{ \cos x}$
log u=cos x.log x
Differentiating w.r.t.x
$\frac{1}{u} \times \frac{du}{dx}=-\sin x \cdot \log x+\cos x \cdot \frac{1}{x}$
$\frac{d u}{dx}=u\left[\frac{\cos x}{x}-\sin x \log x\right]$
$\frac{d y}{d x}=x^{\cos x}\left[\frac{\cos x}{x}-\operatorname{sin} x \log x\right]$
Now , $v=(\cos x)^{\sin x}$
Taking log both sides
$\log v=\log(\cos x)^{\sin x}$
log v=sin x.log(cos x)
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{du}{dx}=\cos x \cdot \log (\cos x)+\sin x \times \frac{1}{\cos x} \times(-\sin x)$
$\frac{d v}{dx}=v[\cos x \log (\cos x)-\sin x \tan x]$
$\frac{d v}{d x}=(\cos )^{\sin} [(\cos x \log x(\cos x)-\sin x \tan x]$
∵ y=u+v
Differentiating w.r.t.x
$\frac{d y}{d x}=\frac{d u}{dx}+\frac{d v}{d x}$
$\frac{d y}{dx}=x^{\cos x}\left(\frac{\cos x}{x}-\sin x \log x )+(\cos x)^{\sin x}.[\cos x \log \cos x-\sin x \tan x]\right.$
(viii) $e^{\sin x}+(\tan x)^x$ को x के सापेज्ञ अवकलित करें ।
Sol :
Let y=u+v$=e^{\sin x}+(\tan x)^{x}$
$u=e^{\sin x}$
Differentiating w.r.t.x
$\frac{d u}{d x}=e^{\sin x} \cdot \cos x$
$v=(\tan x)^{x}$
Taking log both sides
$\log u=\log (\tan x)^{2}$
log u=x.log(tan x)
Differentiating w.r.t.x
$\frac{1}{v} \times \frac{d v}{d x}=1 . \log (\tan x)+x \frac{1}{\tan x} \times \sec ^{2} x$
$\frac{d v}{d x}=(\tan x)^{x}[\log (\tan x)+x \operatorname{cosec} x+\sec x]$
∵ y=u+v
Differentiating w.r.t.x
$\frac{d y}{dx}=\frac{d y}{d x}+\frac{d v}{d x}$
$\frac{d y}{dx}=e^{\sin x} \cdot \cos x+(\tan x)^{x}\left[\log {\tan(x-2)+x \text{cosec x} \sec x}\right.$
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