find the value of h'(2) where h(x) = x^2/f(x), f(2)=3 and f ' (2)= -2

please show all calc involved

Question

find the value of h'(2) where h(x) = x^2/f(x), f(2)=3 and f ' (2)= -2

please show all calc involved

accessdenied · Answer

So, how would you find h'(x) ?

anonymous · Answer

would you plug in 2? no?

anonymous · Answer

write down the quotient rule, then make the replacements

anonymous · Answer

$$(\frac{g}{f})'=\frac{fg'-gf'}{f^2}$$ with 
$$f(x)=f(x), f'(x)=f'(x), g(x)=x^2, g'(x)=2x$$

anonymous · Answer

is this clear? you should get 
$$h'(x)=\frac{2xf(x)-x^2f'(x)}{f^2(x)}$$ then replace x by 2 to get your answer

anonymous · Answer

should i get 20/9

anonymous · Answer

i didn't do it.
the 9 is right

anonymous · Answer

numerator is 
$$2\times 2\times 3+2^2\times 2$$

anonymous · Answer

yeah it is 20

anonymous · Answer

why isnt f^2(x) in the denominator 3^2 (2) which would be 18

anonymous · Answer

$$f(2)=3$$
$$f^2(2)=3^2=9$$

anonymous · Answer

ok i got it. for a moment it looked like f^2 times (x)