Answer to Question #251474 in Calculus for Jennifer

Question

Differentiate from first principle y=tanx

Answer 1

$$(\tan x)'=\lim\limits_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{\dfrac{\sin(x+h)}{\cos(x+h)}-\dfrac{\sin x}{\cos x}}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{\dfrac{\sin(x+h)\cos x-\sin x\cos(x+h)}{\cos(x+h)\cos x}}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{\sin(x+h-x)}{h\cos(x+h)\cos x}$$

$$=\lim\limits_{h\to 0}\dfrac{\sin(h)}{h}\lim\limits_{h\to 0}\dfrac{1}{\cos(x+h)\cos x}$$

$$=1(\dfrac{1}{\cos(x+0)\cos x})$$

$$=\dfrac{1}{\cos^2 x}$$

$$(\tan x)'=\dfrac{1}{\cos^2 x}$$