And yet another one - Chain Rule

[Marcia]

Suppose that f, g are differentiable functions and g(1)=3 , g'(1)=-4 , f'(3)=5.

Let h(x)=f(g(x)). Find h'(1)


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

Is this even a start?

 

[soroban]

Hello, Marcia!

Suppose that f, g are differentiable functions and g(1) = 3 , g'(1) = -4 , f '(3) = 5.

Let h(x) = f(g(x)). Find h'(1)

h'(x) = f(x)g'(x) + g(x)f'(x) . . . Is this even a start?
. . . . . sorry, no . . . it's not a product.

The function is a composite: .h(x) .= .f(g(x)) . . . f of g(x)

We use the Chain Rule: . h'(x) .= .f '(g(x))·g'(x)

Now we want: .h'(1) .= .f '(g(1))·g'(1) . . . . we're told: .g(1) = 3, g'(1) = -4
. . . . . . . . . . . . . . . . . . . . . . .↓. . . . ↓
. . . . . . . . . . . . .h'(1) .= .. f '(3) · (-4) . . . . we're told: .f '(3) = 5
. . . . . . . . . . . . . . . . . . . . . . .↓
. . . . . . . . . . . . .h'(1) .= . . . 5 · (-4) . = . -20