OpenStudy (anonymous):

Calculus help please!

2 years ago
OpenStudy (anonymous):

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).

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

This one involves the product rule :p

2 years ago
OpenStudy (aihberkhan):

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

2 years ago
OpenStudy (aihberkhan):

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

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

So where do I go from there?

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

To find h'(1)

2 years ago
OpenStudy (anonymous):

OH! I get it.

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

@q12157 I got the question right, thank you!

2 years ago