# Thread: Power and Product Rule

1. ## Power and Product Rule

Differentiate the given function (two methods)

I only know how to do the chain rule which I get 8(4x-7)

please show how to do this so I can master the rest of the questions my self..thanks

2. where is your function ??

3. Sorry I can't believe I did not post it:

(7-4x)^2

4. you differentiate it correct there is a negative sign

$\displaystyle f(x)=(7-4x)^2$

$\displaystyle f'(x)=2(-4)(7-4x)$

$\displaystyle f(x)=(-3x^2)^{n}....f'(x)=n(-3x^2)'(-3x^2)^{n-1}=n(-3(2)(x)^{2-1})(-3x^2)^{n-1}$

5. How do I show that the two answers are the same (product rule and power rule) Thanks

6. $\displaystyle f(x)=(7-4x)(7-4x)$

I will differentiate it by product rule

$\displaystyle f'(x)=(7-4x)(-4)+(-4)(7-4x)=2(-4)(7-4x)$

in general

$\displaystyle f(x)=AB...A,B..functions..of..x$

differentiate

$\displaystyle f'(x)=A\frac{d}{dx}(B)+B\frac{d}{dx}(A)$

7. What you did for the power rule..I don't understand it! Is that all the steps for it? I got the product rule that is pretty easy thanks!

8. Originally Posted by lisa1984wilson
What you did for the power rule..I don't understand it! Is that all the steps for it? I got the product rule that is pretty easy thanks!
Originally Posted by Amer
you differentiate it correct there is a negative sign

$\displaystyle f(x)=(7-4x)^2$

$\displaystyle f'(x)=2(-4)(7-4x)$

$\displaystyle f(x)=(-3x^2)^{n}....f'(x)=n(-3x^2)'(-3x^2)^{n-1}=n(-3(2)(x)^{2-1})(-3x^2)^{n-1}$
That is the product rule, the second equation is the differentiated form of your equation, and in the third Amer is showing you how to do it for any equation.

You take your original equation, reduce the power by 1, multiply the whole thing by the original power, this is the $\displaystyle 2$ in your equation, and then multiply the whole thing by the derivative of what's in the brackets, in your equation this is $\displaystyle (-4)$.

Hope this helps

9. I know the product rule its the power rule which I don't understand

10. Originally Posted by lisa1984wilson
I know the product rule its the power rule which I don't understand
If you have an equation $\displaystyle y = (f(x))^n$.

Then $\displaystyle \frac{dy}{dx} = nf'(x)(f(x))^{n-1}$

As I said:

Originally Posted by craig
You take your original equation, reduce the power by 1, multiply the whole thing by the original power, this is the $\displaystyle 2$ in your equation, and then multiply the whole thing by the derivative of what's in the brackets, in your equation this is $\displaystyle (-4)$.

11. sorry I misunderstood...I've been working on calculus all day getting tired...thanks!

12. Originally Posted by lisa1984wilson
I know the product rule its the power rule which I don't understand
$\displaystyle \begin{gathered} \frac{{df^n }} {{dx}} = n \cdot f^{n - 1} \cdot f' \hfill \\ \frac{{d\left( {x^3 + x} \right)^7 }} {{dx}} = 7\left( {x^3 + x} \right)^6 \left( {3x^2 + 1} \right) \hfill \\ \end{gathered}$

13. Originally Posted by lisa1984wilson
sorry I misunderstood...I've been working on calculus all day getting tired...thanks!
No worries, hope it's all cleared up now