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

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.