The functions f(x) and g(x) are differentiable. The function h(x) is defined as:
$h(x)$= $\dfrac{g(x)}{f(x)}$
If f(0)=-7, f′(0)=5, g(0)=3 and g′(0)=–6, find value of h′(0)?

The correct answer is $\dfrac{27}{49}$

Given h(x)= $\dfrac{g(x)}{f(x)}$ and that the functions f(x) and g(x) are differentiable.

To find h′(0), let us apply the quotient rule to find the formula for h′(x).
h′(x) = $\dfrac{g(x)}{f(x)}′$ = $\dfrac{g′(x)f(x)–g(x)f′(x)}{[f(x)]^2}$


Plug in x=0 and use the values f(0)=–7, f′(0)=5, g(0)=3 and g′(0)=–6.

h′(0) = $\dfrac{g′(0)f(0)–g(0)f′(0)}{[f(0)]^2}$ = $\dfrac{–6(–7)–3(5)}{(–7)^2}$ = $\dfrac{27}{49}$

So, h′(0)= $\dfrac{27}{49}$