# Math Help - [SOLVED] Help with derivatives!

1. ## [SOLVED] Help with derivatives!

I am trying to figure this thing out. I am supposed to get the derivative using the product rule, the equation is d/dx * (f(x)^4). I have gotten it down to f(x) * d/dx * (f(x)^3). I am completely stuck. Could anyone help point me in the right direction?

2. I'm going to be a little picky here because I think it will help with your understanding. You don't really have an equation here, as an equation needs what? That's right, an equal sign, =.

The terminology d/dx is meant to indicate that the next term is to be differentiated with respect to x.

Do you have written $f'(x)*f^4(x)$ OR $\frac{d}{dx} f^4(x)$?

3. The 2nd equation is what I have written. I think it is more of a proof than an equation. Please, be as picky as you would like. It will help me to learn it better.

4. Ok good I'm glad you see it's only to help.

So if we have some something f(x) raised to the fourth power, we know that this is a sum of terms of various powers of x, meaning, the standard power rule can apply.

So...

$\frac{d}{dx} f^4(x)=4f^3(x)*f'(x)$

Reduce the power by one and multiply by the derivative of the inner expression.

Got it?

5. I think iv'e got it. So if I had d/dx*f^5x, would it equal 5f^4x * f ' (x)?

6. i thought the product rule was simply

u * dv + v * du

^ if this is right you take your two terms( the terms from the problem in the book). make one of them u and the other v.

so u multiply the u times the derivative of the v. then u add it to v times the derivative of u. simple as that *i think*

7. Originally Posted by UNTEng10
I think iv'e got it. So if I had d/dx*f^5x, would it equal 5f^4x * f ' (x)?
Yes.

Don't write the $\frac{d}{dx}$ TIMES an expression because you're not multiplying them, you're taking d/dx OF an expression.

8. Originally Posted by Legendsn3verdie
i thought the product rule was simply

u * dv + v * du

^ if this is right you take your two terms( the terms from the problem in the book). make one of them u and the other v.

so u multiply the u times the derivative of the v. then u add it to v times the derivative of u. simple as that *i think*

Yes you are right this is the product rule. I realize the OP said to use it but I don't think it's the best way.

Imagine $f(x)=1+x+x^2$.

Thus $(f(x))^4=(1+x+x^2)^4$ and the way I showed holds as the way to calculate the derivative. Do you see that I can keep adding any degree of x and this still holds?

9. Originally Posted by Jameson
Yes.

Don't write the $\frac{d}{dx}$ TIMES an expression because you're not multiplying them, you're taking d/dx OF an expression.
Awesome, thanks for all the help!

10. Originally Posted by Jameson
Yes you are right this is the product rule. I realize the OP said to use it but I don't think it's the best way.

Imagine $f(x)=1+x+x^2$.

Thus $f^4(x)=(1+x+x^2)^4$ and the way I showed holds as the way to calculate the derivative. Do you see that I can keep adding any degree of x and this still holds?

i kinda see but x^2+x+1 doesnt factor hmm.

11. The chain rule says basically that if you have two functions, f(x) and g(x), then $[f(g(x))]' = f'(g(x))*g'(x)$ So let g(x) just be some function g(x) and let f(x)=x^4. Now $f(g(x)) = g^4(x)$ Right?

So f' by itself is obviously 4x^3, now just let x be g(x) and we get the composite derivative is $[4g(x)]^3*g'(x)$

Make sense?

12. Originally Posted by Jameson
The chain rule says basically that if you have two functions, f(x) and g(x), then $[f(g(x))]' = f'(g(x))*g'(x)$ So let g(x) just be some function g(x) and let f(x)=x^4. Now $f(g(x)) = g^4(x)$ Right?

So f' by itself is obviously 4x^3, now just let x be g(x) and we get the composite derivative is $[4g(x)]^3*g'(x)$

Make sense?

yes, the chain rule, the stuff i forgot about the chain rule but now i remember! thanks!

13. Thanks for the help guys, made my life a lot easier! Thanks sent out!