find the derivative?

Question

f(x)= ln(e^-x^20

and

f(theta) = sqrt(sin2Theta)

f(x)= ln(e^-x^2) .. sorry

Answer 1

A) -x^2

B) cos2theta^(-1/2) or

1 / sqrt(cos2theta)

Answer 2

(a)

If y = ln{e^(-x^2)}
then y = -x^2 since ln(e) =1
derivative is -2*x

(b)
derivative w.r.t. theta is
{1/[2*sqrt(sin(2theta) )]} * (cos(2theta) ) *2
cancelling the 2 we get

(cos(2theta) ) / sqrt(sin(2theta) )

Answer 3

Is that f(x)=ln(e^(-x^20))?

If so, then f(x)=(-x^20)• ln(3)= -x^20. Thus f'(x)=-20x^19

f(x)=√(sin(2x)) = (sin(2x))^(1/2)

Thus by chain rule:
f'(x)=1/2(sin(2x))^(-1/2) • cos(2x)•2 = cos(2x)/√(sin(2x))

Answer 4

f(x) = ln (e ^- x²)
f(x) = -x²
f'(x) = -2x
*********************
f(θ) = √sin2θ
f(θ) = (sin2θ) ^ ½
f'(θ) = 1/2 * (sin2θ) ^ -½ * (2cos2θ)
f'(x) = cos2θ / √sin2θ