How do you differentiate $f (x) = tan x + cot x$ $f(x)=tanx+cotx$ ?

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Answer 1 · 2016-09-26T14:38:51

$\frac{d y}{d x} = {sec}^{2} x + (- {csc}^{2} x)$ $dy/dx = sec^(2)x+(-csc^(2)x)$

The derivative is distributive over addition and subtraction.

Hence:

$\frac{d y}{d x} = D_{x} [tan x] + D_{x} [cot x]$ $dy/dx = D_x[tanx] + D_x[cotx]$

Differentiating we get:

$\frac{d y}{d x} = {sec}^{2} x + (- {csc}^{2} x)$ $dy/dx = sec^(2)x+(-csc^(2)x)$

It is important to know the derivatives of the trigonometric functions.

How do you differentiate f(x)=tanx+cotxf(x)=tanx+cotx?