Given f(x) = y, derive y^4, y^9 (Am I using Chain Rule right?)

[mikexz]

Just a quick question

if I have a function f(x)=y
and the question is to derive
1. y^4
2. y^9

My question is in my solution below, am I using the chain rule incorrectly?
1. I know that the answer should be 4(y^3)(dy/dx)

but when I use the chain rule I get this
Steps:
4(y^3) <-- take derivative of outside times derivative of inside
12(y^2)
24(dy/dx)

I know my solution isn't correct since if I derive x^4 it's 4x^3 NOT 4(3)(2)dx/dx

what am I doing wrong? I think I might have misunderstood the concept but I am not sure where.

 

[stapel]

...when I use the chain rule I get this
Steps:
4(y^3) <-- take derivative of outside times derivative of inside

Where is your derivative of the "inside" (by which I assume you mean "f(x)")?

12(y^2)
24(dy/dx)

It looks as though, instead of working your way down through the layers, differentiating as you went, you just kept differentiating the bit with the "y", until you got down to "dy/dx", and then quit. But that's not the Chain Rule. :shock:

To learn how to use the Chain Rule, try this "Ask Dr. Math" article. :wink:

Eliz.

 

[skeeter]

Re: Given f(x) = y, derive y^4, y^9 (Am I using Chain Rule right

consider the function ...
y = (x[sup:23ldqe2w]2[/sup:23ldqe2w] + 1)[sup:23ldqe2w]4[/sup:23ldqe2w]

using the chain rule, you get the derivative

dy/dx = 4(x[sup:23ldqe2w]2[/sup:23ldqe2w] + 1)[sup:23ldqe2w]3[/sup:23ldqe2w](2x)

in other words, you started with (function)[sup:23ldqe2w]4[/sup:23ldqe2w], and took the derivative using the chain rule ... 4(function)[sup:23ldqe2w]3[/sup:23ldqe2w](derivative of function)

so, y is a function ... you have to treat it as such when taking a derivative.

d/dx[y[sup:23ldqe2w]4[/sup:23ldqe2w]] = 4y[sup:23ldqe2w]3[/sup:23ldqe2w](dy/dx) ... dy/dx, the derivative of the function y, is there because of the chain rule, just as shown above.

get the idea?

 

[mikexz]

Re: Given f(x) = y, derive y^4, y^9 (Am I using Chain Rule right

I got it. thanks