What's f'(1)?

Question

If f(x)=(x^2+1)^(2-3x), what's f'(1)?

 

f'(x)=-3*ln(x^2+1)*(x^2+1)^(2-3x)

f'(1)=-3*ln(2)*2^(-1)

 

But the answer is (-1/2)*ln(8e)

 

 

Answer 1

You can change it to e function.

f(x) = e^[(2-3x)ln(x^2+1)]

f'(x) = e^[(2-3x)ln(x^2+1)] [-3ln(x^2+1) + (2-3x)2x/(x^2+1)]

f'(1) = e^[(-1)ln2] [-3ln2 - 1] = -(1/2)ln(8e)

Answer 2

f(x)=(x^2+1)^(2-3x), what's f'(1) Substitute 1 for x f(1) = (1^2 + 1)^(2 - 3_ = (1/2) f(x)=-3*ln(x^2+1)*(x^2+1)^(2-3x) f(1) = (-3)ln(2)*(2)^(-1) = [(-3)ln(2)]/2 This can be rewritten as -(1/2)ln (2^3) or -(1/2)ln(8)