Question

$diffrentitate \frac{ x^5-cosx}{sinx} \: with \: respect \: to \: x$
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Answer 1

ANSWER:

To Do:

• Differentiate (x^5-cosx)/(sinx) wrt x

Solution:

$\text{We are given that,}\\\\:\implies y=\dfrac{x^5-\cos x}{\sin x}\\\\:\implies \dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{x^5-\cos x}{\sin x}\right)\\\\\text{We know that,}\\\\\text{If,}\\\\:\longrightarrow y=\dfrac{u}{v}\\\\\text{Then,}\\\\:\longrightarrow\dfrac{dy}{dx}=\left(\dfrac{u}{v}\right)'\\\\\text{Using Quotient Rule,}\\\\:\hookrightarrow\dfrac{dy}{dx}=\dfrac{vu'-uv'}{v^2}\\\\\text{Here, u = ( x^5-\cos x ) and v = ( \sin x ).}$

$:\implies\dfrac{dy}{dx}=\dfrac{(\sin x)(x^5-\cos x)'-(x^5-\cos x)(\sin x)'}{(\sin x)^2}\\\\\text{We know that,}\\\\:\hookrightarrow \dfrac{d}{dx}x^n=nx^{n-1}\\\\:\hookrightarrow\dfrac{d}{dx}\cos x=-\sin x\\\\:\hookrightarrow\dfrac{d}{dx}\sin x=\cos x\\\\\text{So,}\\\\:\implies\dfrac{dy}{dx}=\dfrac{(\sin x)(x^5-\cos x)'-(x^5-\cos x)(\sin x)'}{(\sin^2x}\\\\:\implies\dfrac{dy}{dx}=\dfrac{(\sin x)[5x^{5-1}-(-\sin x)]-(x^5-\cos x)(\cos x)}{\sin^2x}\\\\:\implies\dfrac{dy}{dx}=\dfrac{(\sin x)[5x^4+\sin x]-(x^5-\cos x)(\cos x)}{\sin^2x}$

$:\implies\dfrac{dy}{dx}=\dfrac{5x^4\sin x+\sin^2x-(x^5\cos x-\cos^2x)}{\sin^2x}\\\\:\implies\dfrac{dy}{dx}=\dfrac{5x^4\sin x+\sin^2x-x^5\cos x+\cos^2x}{\sin^2x}\\\\:\implies\dfrac{dy}{dx}=\dfrac{5x^4\sin x-x^5\cos x+(\sin^2x+\cos^2x)}{\sin^2x}\\\\\text{We know that,}\\\\:\hookrightarrow\sin^2\theta+\cos^2\theta=1\\\\\text{Hence,}\\\\:\implies\dfrac{dy}{dx}=\dfrac{5x^4\sin x-x^5\cos x+1}{\sin^2x}\\\\\bf{:\implies\dfrac{dy}{dx}=\dfrac{-x^5\cos x-5x^4\sin x+1}{\sin^2x}}$

Formulae Used:

$:\hookrightarrow1)\:\text{Quotient Rule(y = u/v)}:\dfrac{dy}{dx}=\dfrac{vu'-uv'}{v^2}\\\\:\hookrightarrow2)\:\dfrac{d}{dx}x^n=nx^{n-1}\\\\:\hookrightarrow3)\:\dfrac{d}{dx}\cos x=-\sin x\\\\:\hookrightarrow4)\:\dfrac{d}{dx}\sin x=\cos x\\\\:\hookrightarrow5)\:\sin^2\theta+\cos^2\theta=1$

Answer 2

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