Find the derivative of the trigonometric function.

Question

Answer 1

Let u= cos(tanπx) then du = -sin(tanπx)πsec²(πx). Then y=sin(u^1/2) and y' = cos(u^1/2)du/2u^1/2. This now gives y' = cos√(cos(tanπx))(-π/2)(sin(tanπx)sec²(πx)/√(cos(tanπx)). This can be simplified to the following form y'= cos√(cos(tanπx))(-π/2)(√(cos(tanπx))tan(tanπx)sec²(πx))