Respuesta :

Hagrid
Remember the chain rule.
L(x)=f(g(x))

L'(x)=f'(g(x))g'(x)

take the derivative of f(g(x)). just treat them like they are variables. so you get:

h'=f'(g(x))g'(x)

now plug in your x value and evaluate:

h'(1)=f'(g(1))(g'(1))


substitute in values that you know and evaluate again
h'(1)=f'(3)(-3)

h'(1)=(-5)(-3)=15