Find the derivative of the trigonometric function.

Question

Answer 1

y^' = πsec²πx·(-sin(tanπx))·1/(2√cos(tanπx))·cos√costanπx

Answer 2

you can write the equation in another way: sin(cos(tan(πx)))^1/2.

from there on, apply the chain rule starting from inside to outside; starting from inside the brackets.

So, derivative of tanπx=πsec²(πx)

derivative of cos(tan(xπ))=-sin(tan(πx))

derivative of sin(cos(tan(πx)))^1/2=1/2(sin(cos(tan(πx)))^-1/2(cos(cos(tan(πx)))

combining all derivatives, we get=

1/2(sin(cos(tan(πx)))^-1/2(cos(cos(tan(πx))) (-sin(tan(πx))) (πsec²(πx))