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Q.Find the derivative of 'tan x' from first principle?

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2024Subjective· 4mImportance★★★★★
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By first principles, the derivative of tan⁡x\tan x is sec⁡2x\sec^2x.

By definition, f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0tan⁡(x+h)−tan⁡xhf'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to0}\dfrac{\tan(x+h)-\tan x}{h}.

Write tan⁡(x+h)−tan⁡x=sin⁡(x+h)cos⁡(x+h)−sin⁡xcos⁡x=sin⁡(x+h)cos⁡x−cos⁡(x+h)sin⁡xcos⁡(x+h)cos⁡x\tan(x+h)-\tan x=\dfrac{\sin(x+h)}{\cos(x+h)}-\dfrac{\sin x}{\cos x}=\dfrac{\sin(x+h)\cos x-\cos(x+h)\sin x}{\cos(x+h)\cos x}.

The numerator is exactly sin⁡((x+h)−x)=sin⁡h\sin((x+h)-x)=\sin h by the sine-difference identity, so:

tan⁡(x+h)−tan⁡x=sin⁡hcos⁡(x+h)cos⁡x\tan(x+h)-\tan x=\dfrac{\sin h}{\cos(x+h)\cos x}

Therefore:

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