Question

find dy/dx if y=cosx÷logx by division rule
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Answer 1

$\bf \large\underbrace\orange{Question:}$

• Find dy/dx if y = cosx ÷ logx by division rule.

$\bf \large \ \underbrace\orange{Answer:}$

Given :-

$\bf \:y = \dfrac{cosx}{logx}$

To find :-

$\bf \:\dfrac{dy}{dx}$

Formula used :-

Division Rule

$\bf \:\dfrac{d}{dx} (\dfrac{u}{v} ) = \dfrac{v\bf \:\dfrac{d}{dx} u - u\bf \:\dfrac{d}{dx} v}{ {v}^{2} }$

$\bf \:\dfrac{d}{dx} logx = \dfrac{1}{x}$

$\bf \:\dfrac{d}{dx} cosx = - sinx$

$\bf\underbrace\red{Solution:}$

$\bf \:y = \dfrac{cosx}{logx}$

Differentiate w. r. t. x, we get

$\bf \:\dfrac{d}{dx} y = \dfrac{d}{dx} \dfrac{cosx}{logx}$

$\bf\implies \: \dfrac{dy}{dx} = \dfrac{logx\bf \:\dfrac{d}{dx} cosx - cosx\bf \:\dfrac{d}{dx} logx}{ {(logx)}^{2} }$

$\bf\implies \: \dfrac{dy}{dx} = \dfrac{logx( - sinx) - cosx \dfrac{1}{x} }{ {(logx)}^{2} }$

$\bf\implies \: \dfrac{dy}{dx} = - \dfrac{xlogx \: sinx + cosx}{ x{(logx)}^{2} }$

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