Question

Find dy/dx if x^4+2x^2y^3+y^2=21

Answer 1

Take the derivative of the function, but wherever you see a y, multiply it by \(\frac{dy}{dx}\) or \(y'\).

Answer 2

\[\frac{dy}{dx}(x^4+2x^2y^3+y^2=21)\]Focus on the \(2x^2y^3\). Use the product rule here, which states \(f'g+g'f\), \(f(x) = 2x^2\) and \(g(x) = y^3\).

Answer 3

Quick question: For 2x^2y^3, can we separate like this, 2(x^2y^3)?

Answer 4

No.

Answer 5

Two separate functions.

Answer 6

Why?

Answer 7

Because the 2 is multiplying the x, and not the y. if it were multiplying both it would read as \(2(x^2y^3)\) in which we would use the product rule first, then multiply in the 2 after.

Answer 8

Oh okay. That's the reason why I got the question wrong. I got it now. Thank you :) @Jhannybean

Answer 9

You figured out the rest then?

Answer 10

Yes

Answer 11

Awesome \(\checkmark\)

Answer 12

Well I hate to be _that guy_ but you can perfectly fine separate it like that 2(x^2y^3) \[\Large \frac{d}{dx}(2*(x^2y^3) )=\frac{d}{dx}(2)*(x^2y^3)+2*\frac{d}{dx}(x^2y^3)\] Now we know that the derivative of 2 with respect to x is 0 cause it's a constant so that first term disappears and we're left with \[\Large 2\frac{d}{dx}(x^2y^3)\] and you can continue to do the product rule on this term. So now you can see why constants can come out of the derivative. =)

Answer 13

If you don't mind, can you show me your work because I did that way and I kept getting the wrong answer...