How do you take the derivative of $cos (tan x)$ $cos(tan x)$ ?

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Answer 1 · 2018-04-18T00:32:00

$- sin (tan (x)) {sec}^{2} (x)$ $-sin(tan(x))sec^2(x)$

Use the chain rule, which states that,

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$

Let $u=tan(x),:.(du)/dx=sec^2(x)$

Then $y=cos(u),:.dy/(du)=-sin(u)$

Combining, we get:

$dy/dx=-sin(u)*sec^2(x)$

$=-sin(u)sec^2(x)$

Substitute back $u=tan(x)$ to get the final answer:

$=-sin(tan(x))sec^2(x)$

How do you take the derivative of cos(tan x)cos(tanx)cos(tan x)?