Find f ' (2) for Simple Derivatives: g(2)=3, g ' (2)=-2, h(2)=-1, h ' (2)=4"

[physicsguy98]

Find f ' (2) given the following.
g(2) = 3 , g ' (2) = -2
h(2) = -1 , h ' (2) = 4

a. f(x) = 2g(x) + h(x)
b. f(x) = g(x) / h(x)
c. f(x) = 4 - h(x)
d. f(x) = g(x)*h(x)

 

[Mark44]

Find f ' (2) given the following.
g(2) = 3 , g ' (2) = -2
h(2) = -1 , h ' (2) = 4

a. f(x) = 2g(x) + h(x)
b. f(x) = g(x) / h(x)
c. f(x) = 4 - h(x)
d. f(x) = g(x)*h(x)

What have you tried. You have to show some work before we are allowed to give you help.

 

[physicsguy98]

I tried using lim(x->c) of f(x) - f(c) / x-c but i end up with lim (x->c) of (2g(x) + h(x) -5)/(x-2) for part (a) and i don't know where to go from there

 

[Mark44]

Don't you know any differentiation rules other than the limit definition? For example, the constant multiple rule, the sum rule, the product rule, the quotient rule?

The limit definition of the derivative can be used, but it will be a pain for products, and especially quotients.