Question

if y = log x/ sin x . find dy / dx​

Answer 1

Step-by-step explanation:

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Answer 2

$\displaystyle \sf{ \frac{dy}{dx} = \frac{sinx - x \: cosx \: logx}{x {sin}^{2} x} }$

Given :

$\displaystyle \sf{ y = \frac{logx}{sinx} }$

To find :

$\displaystyle \sf{ \frac{dy}{dx} }$

Formula :

$\displaystyle \sf{ \frac{d}{dx}\bigg( \frac{u}{v} \bigg) = \frac{v \dfrac{du}{dx} - u \dfrac{dv}{dx} }{ {v}^{2} } }$

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf{ y = \frac{logx}{sinx} }$

Step 2 of 2 :

Find the value of dy/dx

$\displaystyle \sf{ y = \frac{logx}{sinx} }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = \frac{d}{dx}\bigg( \frac{logx}{sinx} \bigg) }$

$\displaystyle \sf \implies \frac{dy}{dx} = \frac{sinx. \frac{d}{dx} (logx) - logx. \frac{d}{dx} (sinx)}{ {(sinx)}^{2} }$

$\displaystyle \sf \implies \frac{dy}{dx} = \frac{sinx. \frac{1}{x} - logx. cosx}{ {sin}^{2} x}$

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{sinx - x \: cosx \: logx}{x {sin}^{2} x}}$

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