# Chain rule question-Is this correct?

• Apr 3rd 2010, 06:41 AM
Neverquit
Chain rule question-Is this correct?
Could someone please double check my workings.

I have this equation
Y=[4th root(x^2+4)]^3

Rearranged is [(x^2+4)^1/4]^3

When I apply the chain rule
Y=[(x^2+4)^1/4]^3

U=(x^2+4)^1/4..
u=1/4(x^2+4)^-3/4.2x

Y=u^3..
Y=3u^2

=u.y3[1/4(x^2+4)^-3/4.2x]^2
=.. 3[1/2x(x^2+4)^-3/4]^2
=... [3/2x(x^2+4)^-3/4]^2
=3x/2(x^2+4)^(-3/4+2/4)]
=3x/2(x^2+4)^-1/4

Solution is 3x/[2(x^2+4)^1/4]

Is my workings correct?
• Apr 3rd 2010, 06:47 AM
Sudharaka
Quote:

Originally Posted by Neverquit
Could someone please double check my workings.

I have this equation
Y=[4th root(x^2+4)]^3

Rearranged is [(x^2+4)^1/4]^3

When I apply the chain rule
Y=[(x^2+4)^1/4]^3

U=(x^2+4)^1/4..
u=1/4(x^2+4)^-3/4.2x

Y=u^3..
Y=3u^2

=u.y3[1/4(x^2+4)^-3/4.2x]^2
=.. 3[1/2x(x^2+4)^-3/4]^2
=... [3/2x(x^2+4)^-3/4]^2
=3x/2(x^2+4)^(-3/4+2/4)]
=3x/2(x^2+4)^-1/4

Solution is 3x/[2(x^2+4)^1/4]

Is my workings correct?

Dear Neverquit,

Your final answer, $\displaystyle \frac{3x}{2(x^2+4)^{\frac{1}{4}}}$ is correct.