Solve the equation below:

Let f and g be differentiable functions such that

f (1) = 4, g (1) = 3, f' (3) = - 5, f' (1) = - 4, g' (1) = - 3, g' (3) = 2

If H (x) = f (g (x)), then h' (1) =

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