# Thread: proving that a derivative is equal to a function

1. ## proving that a derivative is equal to a function

I was given this problem in my calculus class and I have ablsolutely no idea how to solve it. Any help would be greatly appreciated.

Consider the two functions F and G whose derivates Fprime and Gprime exist. the functions have the properties listed below for all values of x. Prove Gprime(x)=-F(x)
I. (F(x))^2 + (G(x))^2=1
II. Fprime(x)=G(x)
III. G(x)>0
IV. F(0)=0

2. Originally Posted by wisezeta
I was given this problem in my calculus class and I have ablsolutely no idea how to solve it. Any help would be greatly appreciated.

Consider the two functions F and G whose derivates Fprime and Gprime exist. the functions have the properties listed below for all values of x. Prove Gprime(x)=-F(x)
I. (F(x))^2 + (G(x))^2=1
II. Fprime(x)=G(x)
III. G(x)>0
IV. F(0)=0
Differentiate I. You get 2F*F'+2G*G'=0. Use II to substitute: 2F*G+2G*G'=0. From there simplifying will show your intended relation. I'm trying to think of if you need III or IV to justify anything, but for now that is the basic layout of the proof.

3. Try taking the derivative:

$\displaystyle \frac{d}{dx}(F(x)^2+G(x)^2)=\frac{d}{dx}1$

$\displaystyle 2F(x)F'(x)+2G(x)G'(x)=0$

Now supstitute $\displaystyle F'(x)=G(x)$

You will need (iii) to justify division by G(x)

4. Originally Posted by hjortur

You will need (iii) to justify division by G(x)
Isn't all you need to have $\displaystyle G(x) \ne 0$ for the division?

5. Yes, and G(x)>0 says that G(x)!=0.

But I still cant see what F(0)=0 has to do with anything.