Suppose that f and g are functions that are differentiable at x = 1 and that f(1) = 4, f '(1) = -3, g(1) = -4, and g'(1) = 5. Find h'(1). h(x)=f(x)g(x)/f(x)-g(x)

Answer 1

By applying the quotient rule, you can evaluate the derivative of h(x) with respect to x and you will obtain the following

h'(x)=Numerator/Denominator

where

Numerator=[ f'(x)·g(x)+f(x)·g'(x) ]*(f(x)-g(x)) - f(x)·g(x)( f'(x)-g'(x) )

Denominator= (f(x)-g(x))²

Since f(1)=4, f'(1)=-3, g(1)=-4, g'(1)=5 we have

Numerator=[-3*(-4) + 4*5]*(4-(-4))+16*(-3-5)=(12+20)*8+16*(-8)=128

Denominator=64

Hence you have h'(1)=128/64=2

Best,

Davide