Question

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) = the quotient of f of x and g of x.

Answer 1

Answer:

h'(1)=0

Step-by-step explanation:

We use the definition of the derivative of a quotient:

If $h(x)=\frac{f(x)}{g(x)}$, then:

$h'(x)=\frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}$

Since in our case we want the derivative of $h(x)$ at the point x = 1, which is indicated by: h'(1), we need to evaluate the previous expression at x = 1, that is:

$h'(1)=\frac{f'(1)*g(1)-f(1)*g'(1)}{(g(1))^2}$

which, by replacing with the given numerical values:

$f(1) =4\\g(1)=3\\f'(1)=-4\\g'(1)=-3$

becomes:

$h'(1)=\frac{f'(1)*g(1)-f(1)*g'(1)}{(g(1))^2}=\\=\frac{-4*3-4*(-3)}{(3)^2}=\frac{-12+12}{9} =\frac{0}{9} =0$