Question

Calculus help please!

Answer 1

If f and g are differentiable functions for all real values of x such that f(1) = 4, g(1) = 3, f '(3) = −5, f '(1) = −4, g '(1) = −3, g '(3) = 2, then find h '(1) if h(x) = f(x) g(x).

Answer 2

I've done the other questions, but I can barely start on this one.

Answer 3

This one involves the product rule :p

Answer 4

Okay.

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

Answer 5

Hope this helped! Have a great day!

Also a medal would be much appreciated! Just click best response next to my answer. Thank You!

@epistemeal

Answer 6

Actually what the person above said is correct if you are doing f(g(x)). If you are doing f(x)*g(x) you need to do product rule

Answer 7

That says that f'(x) *g(x) = dy/dx f(x)*d(x) or no?

Answer 8

the derivative of f(x)*g(x) in other words h'(x) is equal to f'(x)*g(x)+g'(x)*f(x)

Answer 9

So where do I go from there?

Answer 10

well since by the definition of product rule

d/dx( f(x)*g(x) ) = f'(x)*g(x)+g'(x)*f(x)

and h'(x) = d/dx( f(x)*g(x) )

that means

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

Using this you can just plug in numbers

Answer 11

To find h'(1)

Answer 12

OH! I get it.

Answer 13

yep yep! The person above me was explaining f(g(x)) which is the chain rule instead of the product rule lol.

Answer 14

The teacher might be tricky and slip in a quotient rule too but they are all really the same :p

Answer 15

@q12157 I got the question right, thank you!