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) = f(x)/g(x)

Answer 1

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] you have all the number you need, plug them in

Answer 2

By the quotient rule:

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\frac{f(x)}{g(x)}\right] =\frac{f'(x)g(x)-f(x)g'(x)}{\left\{g(x)\right\}^2}}\)

At x=1 (or h(1)), you get:

\(\large\color{black}{ \displaystyle h'(1)=\frac{f'(1)g(1)-f(1)g'(1)}{\left\{g(1)\right\}^2}}\)

Answer 3

then, you are given that:

f(1)=4
g(1)=3

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

Answer 4

Go for it...

Answer 5

so 0?

Answer 6

Yes.

f '(1) g(1) = f(1) g '(1)
(-4) ( 3) = (4) ( -3)
-12=-12

Answer 7

-12-(-12)=-12+12=0

Answer 8

very nice

Answer 9

thanks so much :)

Answer 10

yw