## onegirl 2 years ago Find the derivative with and without using the chain rule. f(x) = (x^2 + 1)^3

1. ingenuus

$f(x)=(x^2+1)^3=(x^2+1)(x^2+1)(x^2+1)$ $f'(x)=6x(x^2+1)^2$

2. ingenuus

used only product rule

3. onegirl

is that using the chain rule?

4. ingenuus

no. i said i only used product rule.

5. onegirl

okay

6. zepdrix

One, do you understand how to find the derivative with using chain rule? You would just take the derivative as is.

7. onegirl

not 100%

8. onegirl

okay i'll try doing with the chain rule

9. zepdrix

Ok :)

10. onegirl

okay can i post it here and correct me if i i mistake somewhere

11. zepdrix

k

12. onegirl

$d/dx ((x^2 + 1)^3) = 3(x^2 + 1)^2 d/dx(x^2 + 1) = 3(d/dx(x^2)(x^2 - 1)^2 = 3(2x)(x^2 = 1)^2 = 6x(x^2 + 1)^2$

13. onegirl

@zepdrix

14. zepdrix

I'm not quite sure why it changed from x^2+1 to x^2-1. Was that a typo?

15. onegirl

i see that it didn't show up all of it but here is the last part 3(2x)(x^2 + 1)^2 = 6x(x^2 + 1)^2

16. onegirl

yes typo for that one

17. zepdrix

$\large f'(x)=3(x^2+1)^2\frac{d}{dx}(x^2+1)$$\large f'(x)=3(x^2+1)^2(2x) \qquad \qquad f'(x)=6x(x^2+1)^2$ Ok yah looks good :)

18. onegirl

thanks!

19. zepdrix

without chain rule is going to be a little tricky. They want you multiply out your binomial :O oh boy.

20. onegirl

ohh so the one @ingenuus showed is nto by without chain rule?

21. onegirl

not*

22. onegirl

the one ingenuus showed its using the product rule

23. zepdrix

That's another method for doing it without chain rule. Which involved the product rule with 3 terms. Which is a little bit tricky if you've never seen it before. Product rule with 2 terms is probably what you're used to seeing. I was referring to a different method, multiplying out all the brackets as Wio showed you the last time you posted this. :D

24. onegirl

Ohhhh okay got you