Question

Its solution is b... Answer only if u know... pls do not spam​

Answer 1

AnsWer :

B)

To FinD :

$\tt \bullet \: \: \: \: \: \dfrac{d}{dx} \bigg \lgroup {x}^{2} .Sin \frac{1}{x} \bigg\rgroup$

SolutioN :

$\tt \bullet \: \: \: \: \: y = {x}^{2} .sin \frac{1}{x}$

Now,

> Differentiation Rule.

• Constant Rule.
• Product Rule.
• Quotient Rule.
• Chain Rule.

Apply Product Rule of Differentiation.

$\tt \longmapsto \: \: \: \: \: \dfrac{dy}{dx} = \dfrac{d}{dy} \bigg \lgroup {x}^{2} sin \frac{1}{x} \bigg\rgroup$

$\tt \mapsto \: \: \: \: \: \dfrac{dy}{dx} = \dfrac{d}{dx} \Big({x}^{2} \Big) .Sin \frac{1}{x}+{x}^{2}. \dfrac{d}{dx} \Big(Sin \frac{1}{x}\Big)$

$\tt \mapsto \: \: \: \: \: \dfrac{dy}{dx} = \dfrac{d}{dx} \Big({x}^{2} \Big) .Sin \frac{1}{x}+{x}^{2}. \dfrac{d}{dx} \Big(Sin \frac{1}{x}. \frac{1}{x} \Big)$

$\tt \mapsto \: \: \: \: \: \dfrac{dy}{dx} = 2x .Sin \frac{1}{x}+{x}^{2}.Cos\frac{1}{x}.\Big( - \frac{1}{ {x}^{2} } \Big)$

$\tt \mapsto \: \: \: \: \: \dfrac{dy}{dx} = 2x .Sin \frac{1}{x} - \cancel{{x}^{2} }. \frac{1}{ \cancel{ {x}^{2} }} .Cos\frac{1}{x}$

$\tt \mapsto \: \: \: \: \: \dfrac{dy}{dx} = 2x .Sin \frac{1}{x} - Cos\frac{1}{x}$

Or,

$\tt \mapsto \: \: \: \: \: \dfrac{dy}{dx} = 2xSin\Big(\frac{1}{x}\Big )- Cos\Big(\frac{1}{x}\Big)$

Therefore, Option B ) Correct.

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MorE InformatioN :

$\tt \bullet \: \: \: \: \: y = Sin^nx.$

$\tt \dfrac{dy}{dx} = \dfrac{d\,\Big(Sin^nx\Big)}{dx}$

$\tt \dfrac{dy}{dx} = \dfrac{d\,\Big(Sin^nx\Big)}{dx}.\dfrac{d\,\Big(x\Big)}{dx}$

$\tt \dfrac{dy}{dx} =n Cos^{n-1}x.1.$

$\tt \dfrac{dy}{dx} = nCos^{n-1}x.$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{\dfrac{d}{dx}\left(x^2sin\:\dfrac{1}{x}\right)}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\dfrac{d}{dx}\left(x^2\sin \left(\dfrac{1}{x}\right)\right)=2x\sin \left(\dfrac{1}{x}\right)-\cos \left(\dfrac{1}{x}\right)}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\mathrm{Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'$

$f=x^2,\:g=\sin \left(\dfrac{1}{x}\right)$

$=\dfrac{d}{dx}\left(x^2\right)\sin \left(\dfrac{1}{x}\right)+\dfrac{d}{dx}\left(\sin \left(\dfrac{1}{x}\right)\right)x^2$

$\sf{Solve\:\:\:\dfrac{d}{dx}\left(x^2\right)}$

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$\text{Apply the Power Rule:} \\\\\dfrac{d}{d x}\left(x^{a}\right)=a \cdot x^{a-1} \\\\ =2 x^{2-1} \\\\Simplify\\\\ =2 x$

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$\sf{Solve\:\:\:\dfrac{d}{dx}\left(\sin \left(\dfrac{1}{x}\right)\right)}$

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$\begin{aligned}&\text { Apply the chain rule: } \cos \left(\frac{1}{x}\right) \frac{d}{d x}\left(\frac{1}{x}\right)\\\\&=\cos \left(\frac{1}{x}\right) \frac{d}{d x}\left(\frac{1}{x}\right)\\\\&\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}\\\\&=\cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^{2}}\right)\\\\&\text { Simplify } \cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^{2}}\right):-\frac{\cos \left(\frac{1}{x}\right)}{x^{2}}\\\\&=-\frac{\cos \left(\frac{1}{x}\right)}{x^{2}}\end{aligned}$

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$\sf{=2x\sin \left(\dfrac{1}{x}\right)+\left(-\dfrac{\cos \left(\dfrac{1}{x}\right)}{x^2}\right)x^2}$

$\boxed{\sf{=2x\sin \left(\dfrac{1}{x}\right)-\cos \left(\dfrac{1}{x}\right)}}$