Compute the derivative of tan x. | vedclass

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(B) Let $f(x) = 	an x$. Then by the definition of the derivative:$\frac{df(x)}{dx} = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$$= \lim_{h 	o 0} \frac

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$\sec^2 x$

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(B) Let $f(x) = 	an x$. Then by the definition of the derivative:$\frac{df(x)}{dx} = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$$= \lim_{h 	o 0} \frac{	an(x + h) - 	an x}{h}$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{\sin(x + h)}{\cos(x + h)} - \frac{\sin x}{\cos x} ight]$$= \lim_{h 	o 0} \frac{\sin(x + h)\cos x - \cos(x + h)\sin x}{h \cos(x + h)\cos x}$Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$= \lim_{h 	o 0} \frac{\sin(x + h - x)}{h \cos(x + h)\cos x}$$= \lim_{h 	o 0} \left( \frac{\sin h}{h} ight) \cdot \lim_{h 	o 0} \left( \frac{1}{\cos(x + h)\cos x} ight)$Since $\lim_{h 	o 0} \frac{\sin h}{h} = 1$:$= 1 \cdot \frac{1}{\cos^2 x} = \sec^2 x$

Compute the derivative of $\tan x$.

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