If g(1) = 4 and g¢ (1) = − 2, find f ¢ (1), where f(x) = x2 ln (g(x)). Also find the equation of the

Question

normal to f at x = 1.

Answer 1

I guess f ¢ means f'. So,

f(x) = x^2 ln(g(x))
f' = 2x ln(g(x)) + x^2/g(x) g'
f'(1) = 2 ln4 + -2/4 = (4ln4 - 1)/2

f(1) = ln4

The line normal to f at x=1 has slope 2/(1 - 4ln4)

So, the line through (1,4ln4) with that slope is

y - 4ln4 = 2/(1 - 4ln4) * (x-1)

Answer 2

That would be

So, the line through (1,ln4) with that slope is

y - ln4 = 2/(1 - 4ln4) * (x-1)