Answer to Question #203446 in Calculus for jay

Question

Show that 𝑑𝑦/𝑑𝑥 = 𝑠𝑒𝑐^2 x given that 𝑦 = tan x

Answer 1

$$\dfrac{d}{dx}(\tan (x))=\lim\limits_{h\to0}\dfrac{\tan(x+h)-\tan(x)}{h}$$

$$=\lim\limits_{h\to0}\dfrac{\sin(x+h)\cos(x)-\cos(x+h)\sin(x)}{h \cos(x+h)\cos(x)}$$

$$=\lim\limits_{h\to0}\dfrac{\sin(x+h-x)}{h \cos(x+h)\cos(x)}$$

$$=\lim\limits_{h\to0}\dfrac{\sin(x+h-x)}{h \cos(x+h)\cos(x)}$$

$$=\lim\limits_{h\to0}\dfrac{\sin(h)}{h }\cdot\lim\limits_{h\to0}\dfrac{1}{ \cos(x+h)\cos(x)}$$

$$=1\cdot\lim\limits_{h\to0}\dfrac{1}{h \cos(x+h)\cos(x)}$$

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

$$=\sec^2 (x)$$

$$\dfrac{d}{dx}(\tan (x))=\sec^2(x)$$