Question

Calculus derivative. How to do? (50 points lol)

Answer 1

The derivative operator distributes over sums, so

`(a f(x) + b g(x))' = a f'(x) + b g'(x)`

Then

`h(x) = 5f(x) - 4g(x) \implies h'(x) = 5 f'(x) - 4 g'(x) \\\\ \implies h'(2) = 5(-2) - 4(7) = \boxed{-38}`

Use the product rule,

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

Then

`h(x) = f(x) g(x) \implies h'(x) = f'(x) g(x) + f(x) g'(x) \\\\ \implies h'(2) = (-2) (4) + (-3) (7) = \boxed{-29}`

Use the quotient rule,

`\left((f(x))/(g(x))\right)' = (f'(x) g(x) - f(x) g'(x))/(g(x)^2)`

Then

`h(x) = (f(x))/(g(x)) \implies h'(x) = (f'(x) g(x) - f(x) g'(x))/(g(x)^2) \\\\ \implies h'(2) = ((-2)(4) - (-3)(7))/(4^2) = \boxed{(13)/(16)}`

Use the quotient rule again.

`h(x) = (g(x))/(1 + f(x)) \implies h'(x) = (g'(x) (1+f(x)) - g(x) f'(x))/((1+f(x))^2) \\\\ \implies h'(2) = (7(1 - 3) - 4 (-2))/((1 - 3)^2) = -\frac64 = \boxed{-\frac32}`