Differentiate y=(x^2+2)(cube root (x^2+2))

Question

e.mccormick · Answer

Looks like product rule and a chain.

anonymous · Answer

i just learn chain rule today so i am a little unsure of where to use it

e.mccormick · Answer

AH.  OK.  Fist, let me write this out in LaTeX.

e.mccormick · Answer

$y=(x^2+2)(\sqrt[3]{x^2+2})$
OK.  for the product rule part you have (u)(v).  That is not too bad.  But when you do the product rule you end up with needing the primes of each.  That is where the chain comes in.

e.mccormick · Answer

The cahin rule states that whan you take a derivative of something that contains a derivative, you do a little multiplication:
$$\frac{dy}{dx}f(g(x))=f'(g(x))g'(x)$$What that means is take the derivative of the outer function, which is the cube root in the right most part of yours, and leave the inner function alone.  Then, multiply that times the derivitive of the iner part.

anonymous · Answer

alright let me try to solve it

e.mccormick · Answer

So when you need v' you need $\frac{dy}{xd}\sqrt[3]{x^2+2}$  Start by rewriting that as a fractional exponent.
$$(\sqrt[3]{x^2+2})'\Rightarrow 
(x^2+2)^{1/3}
$$
Then it should be easer to do that part.

anonymous · Answer

okay so thats where i use the chain rule right

e.mccormick · Answer

Yah.

anonymous · Answer

so for that i got 1/3(x^2+2)^-2/3(2x)

e.mccormick · Answer

And that is for the v' part.  So your uv=u'v+uv' it is just that last v' part.

raden · Answer

hint :
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