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Miscellaneous Exercise · Q18

Q.Find the derivative of sec⁡x−1sec⁡x+1\dfrac{\sec x - 1}{\sec x + 1}.

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Rewrite in terms of cos⁡x\cos x to get 1−cos⁡x1+cos⁡x\dfrac{1-\cos x}{1+\cos x}, then apply the Quotient Rule; the derivative is 2sin⁡x(1+cos⁡x)2\dfrac{2\sin x}{(1+\cos x)^2}, equivalently 2sec⁡xtan⁡x(sec⁡x+1)2\dfrac{2\sec x\tan x}{(\sec x+1)^2}.

Both numerator and denominator contain sec⁡x\sec x, so rewriting everything in terms of cos⁡x\cos x makes the algebra cleaner before differentiating.

Step 1 — Simplify using sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}.

y=sec⁡x−1sec⁡x+1=1cos⁡x−11cos⁡x+1=1−cos⁡xcos⁡x1+cos⁡xcos⁡x=1−cos⁡x1+cos⁡x.y=\frac{\sec x-1}{\sec x+1}=\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}=\frac{\frac{1-\cos x}{\cos x}}{\frac{1+\cos x}{\cos x}}=\frac{1-\cos x}{1+\cos x}.

Step 2 — Apply the Quotient Rule.

Let u=1−cos⁡xu=1-\cos x and v=1+cos⁡xv=1+\cos x, so u′=sin⁡xu'=\sin x and v′=−sin⁡xv'=-\sin x. Then

dydx=u′v−uv′v2=(sin⁡x)(1+cos⁡x)−(1−cos⁡x)(−sin⁡x)(1+cos⁡x)2.\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(\sin x)(1+\cos x)-(1-\cos x)(-\sin x)}{(1+\cos x)^2}.

Step 3 — Expand and simplify the numerator.

(sin⁡x)(1+cos⁡x)+(1−cos⁡x)(sin⁡x)=sin⁡x+sin⁡xcos⁡x+sin⁡x−sin⁡xcos⁡x=2sin⁡x,(\sin x)(1+\cos x)+(1-\cos x)(\sin x)=\sin x+\sin x\cos x+\sin x-\sin x\cos x=2\sin x,

so …

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