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Q.Find the derivative of tan⁡x\tan x from the first principle.

Bihar BsebBihar Board Intermediate 1st Year 2025Subjective· 5mImportance★★★★★
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ddx(tan⁡x)=sec⁡2x\dfrac{d}{dx}(\tan x)=\sec^2x, proved from first principles.

Let f(x)=tan⁡xf(x)=\tan x. By definition:

f′(x)=lim⁡h→0tan⁡(x+h)−tan⁡xhf'(x)=\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 sin⁡((x+h)−x)=sin⁡h\sin\big((x+h)-x\big)=\sin h (using sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B)=\sin A\cos B-\cos A\sin B).

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}.

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