Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
\[f(x)=(x^2+1)^3=(x^2+1)(x^2+1)(x^2+1)\] \[f'(x)=6x(x^2+1)^2\]
used only product rule
is that using the chain rule?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

no. i said i only used product rule.
okay
One, do you understand how to find the derivative `with` using chain rule? You would just take the derivative as is.
not 100%
okay i'll try doing with the chain rule
Ok :)
okay can i post it here and correct me if i i mistake somewhere
k
\[d/dx ((x^2 + 1)^3) = 3(x^2 + 1)^2 d/dx(x^2 + 1) = 3(d/dx(x^2)(x^2 - 1)^2 = 3(2x)(x^2 = 1)^2 = 6x(x^2 + 1)^2 \]
I'm not quite sure why it changed from x^2+1 to x^2-1. Was that a typo?
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
yes typo for that one
\[\large f'(x)=3(x^2+1)^2\frac{d}{dx}(x^2+1)\]\[\large f'(x)=3(x^2+1)^2(2x) \qquad \qquad f'(x)=6x(x^2+1)^2\] Ok yah looks good :)
thanks!
`without` chain rule is going to be a little tricky. They want you multiply out your binomial :O oh boy.
ohh so the one @ingenuus showed is nto by without chain rule?
not*
the one ingenuus showed its using the product rule
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
Ohhhh okay got you

Not the answer you are looking for?

Search for more explanations.

Ask your own question