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

Answer 1

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

Answer 2

would you plug in 2? no?

Answer 3

write down the quotient rule, then make the replacements

Answer 4

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

Answer 5

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

Answer 6

should i get 20/9

Answer 7

i didn't do it.
the 9 is right

Answer 8

numerator is
\[2\times 2\times 3+2^2\times 2\]

Answer 9

yeah it is 20

Answer 10

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

Answer 11

\[f(2)=3\]
\[f^2(2)=3^2=9\]

Answer 12

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