What is the second derivative of y=tan x-sinx?

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We'll have to differentiate twice the function, with respect to x.

dy/dx = d(tanx )/dx - d(sin x)/dx

dy/dx = 1/(cos x)^2 - cos x

dy/dx = [1 - (cos x)^3]/(cos x)^2

We'll differentiate again dy/dx, with respect to x. We'll apply the quotient rule:

y" = d^2y/dx^2 = [1 - (cos x)^3]'*(cos x)^2 - [1 - (cos x)^3]*[(cos x)^2]'/(cos x)^4

y" = {3sin x*(cos x)^4 + 2sin x*cos x* [1 - (cos x)^3]}/(cos x)^4

y" = sin x*cos x*[3(cos x)^3 + 2 - 2(cos x)^3]/(cos x)^4

y" = sin x*[(cos x)^3+2]/(cos x)^3

y" = sin x + 2*tan x/(cos x)^2

**The second derivative of the given function is: y" = sin x + 2*tan x/(cos x)^2.**

The second derivative of y = tan x - sin x is the derivative of the first derivative of y.

y = tan x - sin x

Using the formulae for the derivative of tan x and sin x

y' = (sec x)^2 - cos x

y'' = [(sec x)^2 - cos x]'

use the chain rule and the formula for the derivative of cos x

=> 2*sec x * sec x * tan x + sin x

=> 2*(sec x)^2*tan x + sin x

**The second derivative is 2*(sec x)^2*tan x + sin x**

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