please answer this question


​

Answer:

[tex]\dfrac{dy}{dx}=\cot x+\dfrac{1}{x}[/tex]

Step-by-step explanation:

[tex]\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}}[/tex]

[tex]\textsf{Given}:y=\log (x \sin x)[/tex]

[tex]\textsf{Let }u=x \sin x \implies y=\log(u)[/tex]

Using the product rule, differentiate y with respect to [tex]u[/tex]:

[tex]\textsf{If }y=\log(u) \implies \dfrac{dy}{du}=\dfrac{1}{u}\implies \dfrac{dy}{du}=\dfrac{1}{x \sin x}[/tex]

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Differentiate [tex]u=x \sin x[/tex] with respect to x using the product rule:

[tex]\implies \textsf{Let }f(x)=x \implies f'(x)=1[/tex]

[tex]\implies \textsf{Let }g(x)=\sin x \implies g'(x)=\cos x[/tex]

[tex]\implies \dfrac{du}{dx}=x \cos x + \sin x[/tex]

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

[tex]\implies \dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}[/tex]

[tex]\implies \dfrac{dy}{dx}=\dfrac{1}{x \sin x} \times (x \cos x + \sin x)[/tex]

[tex]\implies \dfrac{dy}{dx}=\dfrac{x \cos x + \sin x}{x \sin x}[/tex]

[tex]\implies \dfrac{dy}{dx}=\dfrac{x \cos x}{x \sin x}+\dfrac{\sin x}{x \sin x}[/tex]

[tex]\implies \dfrac{dy}{dx}=\cot x+\dfrac{1}{x}[/tex]

Answer:

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

Step-by-step explanation:

Given:

[tex]\displaystyle \large{y=\log (x\sin x)}[/tex]

Recall the (common) logarithmic differentiation & chain rules:

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

Let [tex]\displaystyle \large{u=x\sin x}[/tex] then:

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

Recall product rules:

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

Hence:

[tex]\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}}[/tex]

From above, recall that:

[tex]\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}[/tex]

Hence, the answer is:

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

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