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Miscellaneous Examples · Example 22

Q.Find the derivative of

(i) x5−cos⁡xsin⁡x\dfrac{x^5 - \cos x}{\sin x}
(ii) x+cos⁡xtan⁡x\dfrac{x + \cos x}{\tan x}
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Both are quotients, so apply the Quotient Rule (uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}. (i) 5x4sin⁡x−x5cos⁡x+1sin⁡2x\dfrac{5x^4\sin x-x^5\cos x+1}{\sin^2 x};

(ii) (1−sin⁡x)sin⁡xcos⁡x−(x+cos⁡x)sin⁡2x\dfrac{(1-\sin x)\sin x\cos x-(x+\cos x)}{\sin^2 x}.

ddx(u(x)v(x))=u′(x) v(x)−u(x) v′(x)[v(x)]2\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)\,v(x) - u(x)\,v'(x)}{[v(x)]^2}


(i) x5−cos⁡xsin⁡x\dfrac{x^5 - \cos x}{\sin x}

Step 1 — Identify parts. u=x5−cos⁡x, v=sin⁡xu=x^5-\cos x,\ v=\sin x.

Step 2 — Differentiate. u′=5x4+sin⁡x,v′=cos⁡xu'=5x^4+\sin x,\quad v'=\cos x.

Step 3 — Quotient rule.

(5x4+sin⁡x)sin⁡x−(x5−cos⁡x)cos⁡xsin⁡2x\frac{(5x^4+\sin x)\sin x-(x^5-\cos x)\cos x}{\sin^2 x}

Step 4 — Expand the numerator.

5x4sin⁡x+sin⁡2x−x5cos⁡x+cos⁡2x5x^4\sin x+\sin^2 x-x^5\cos x+\cos^2 x

Step 5 — Use sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1.

5x4sin⁡x−x5cos⁡x+15x^4\sin x-x^5\cos x+1

ddx ⁣(x5−cos⁡xsin⁡x)=5x4sin⁡x−x5cos⁡x+1sin⁡2x\boxed{\dfrac{d}{dx}\!\left(\dfrac{x^5-\cos x}{\sin x}\right)=\dfrac{5x^4\sin x-x^5\cos x+1}{\sin^2 x}}


(ii) x+cos⁡xtan⁡x\dfrac{x + \cos x}{\tan x}

Step 1 — Identify parts. u=x+cos⁡x, v=tan⁡xu=x+\cos x,\ v=\tan x.

Step 2 — Differentiate. u′=1−sin⁡x,v′=sec⁡2xu'=1-\sin x,\quad v'=\sec^2 x.

Step 3 — Quotient rule.

(1−sin⁡x)tan⁡x−(x+cos⁡x)sec⁡2xtan⁡2x\frac{(1-\sin x)\tan x-(x+\cos x)\sec^2 x}{\tan^2 x}

Step 4 — Convert to sine/cosine. Multiply numerator and denominator by cos⁡2x\cos^2 x. Since tan⁡xcos⁡2x=sin⁡xcos⁡x\tan x\cos^2 x=\sin x\cos x, sec⁡2xcos⁡2x=1\sec^2 x\cos^2 x=1, and tan⁡2xcos⁡2x=sin⁡2x\tan^2 x\cos^2 x=\sin^2 x: …

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