Question

please answer this question


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Answer 1

Answer:

$\dfrac{dy}{dx}=\cot x+\dfrac{1}{x}$

Step-by-step explanation:

$\boxed{\begin{minipage}{6 cm}\underline{Product\:Rule}\\\\ \dfrac{d}{dx}[f(x)g(x)]=f(x)g'(x)+f'(x)g(x) \\\\\\\underline{Chain\:Rule}\\\\ \dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx} \\\end{minipage}}$

$\textsf{Given}:y=\log (x \sin x)$

$\textsf{Let }u=x \sin x \implies y=\log(u)$

Using the product rule, differentiate y with respect to $u$:

$\textsf{If }y=\log(u) \implies \dfrac{dy}{du}=\dfrac{1}{u}\implies \dfrac{dy}{du}=\dfrac{1}{x \sin x}$

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Differentiate $u=x \sin x$ with respect to x using the product rule:

$\implies \textsf{Let }f(x)=x \implies f'(x)=1$

$\implies \textsf{Let }g(x)=\sin x \implies g'(x)=\cos x$

$\implies \dfrac{du}{dx}=x \cos x + \sin x$

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Use the chain rule to differentiate y with respect to x:

$\implies \dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}$

$\implies \dfrac{dy}{dx}=\dfrac{1}{x \sin x} \times (x \cos x + \sin x)$

$\implies \dfrac{dy}{dx}=\dfrac{x \cos x + \sin x}{x \sin x}$

$\implies \dfrac{dy}{dx}=\dfrac{x \cos x}{x \sin x}+\dfrac{\sin x}{x \sin x}$

$\implies \dfrac{dy}{dx}=\cot x+\dfrac{1}{x}$

Answer 2

Answer:

$\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{1}{x\ln 10} + \dfrac{\cot x}{\ln 10}}$

Step-by-step explanation:

Given:

$\displaystyle \large{y=\log (x\sin x)}$

Recall the (common) logarithmic differentiation & chain rules:

$\displaystyle \large{\dfrac{d}{dx}[\log (u)] = \dfrac{u'}{u\ln 10}}$

Let $\displaystyle \large{u=x\sin x}$ then:

$\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{(x\sin x)'}{x\sin x \ln 10}}$

Recall product rules:

$\displaystyle \large{[f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)}$

Hence:

$\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{(x)'\sin x + x(\sin x)'}{x\sin x \ln 10}}\\\\\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{\sin x + x\cos x}{x\sin x \ln 10}}\\\\\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{\sin x}{x \sin x \ln 10} + \dfrac{x \cos x}{x\sin x \ln 10}}\\\\\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{1}{x\ln 10} + \dfrac{\cot x}{\ln 10}}$

From above, recall that:

$\displaystyle \large{\dfrac{d}{dx}(\sin x) = \cos x}\\\\\displaystyle \large{\dfrac{A\pm B}{C} = \dfrac{A}{C} \pm \dfrac{B}{C}}\\\\\displaystyle \large{\dfrac{\cos x}{\sin x} =\cot x}$

Hence, the answer is:

$\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{1}{x\ln 10} + \dfrac{\cot x}{\ln 10}}$