# Derivative of a function!

• Mar 13th 2007, 02:22 PM
coolaid2u
Derivative of a function!
How would you find the derivative for these three problems?? I end up with different answers each time I do them.

y= (4x^2+7)^2 * (3x^3+1)^4

y= (2x)^(1/2) + (2/x)^(1/2)

y= (x^2-5)^(1/2) * (x^2+3)^(1/3)

I'd be happy if anyone could help! : )
• Mar 13th 2007, 02:42 PM
Jhevon
Quote:

Originally Posted by coolaid2u
How would you find the derivative for these three problems?? I end up with different answers each time I do them.

y= (4x^2+7)^2 * (3x^3+1)^4

y= (2x)^(1/2) + (2/x)^(1/2)

y= (x^2-5)^(1/2) * (x^2+3)^(1/3)

I'd be happy if anyone could help! : )

y= (4x^2+7)^2 * (3x^3+1)^4 ............we need the chain rule and product rule here
=> y' = 2(4x^2 + 7)(8x)(3x^3 + 1)^4 + (4x^2+7)^2 * 4(3x^3 + 1)^3 * 9x^2
=> y' = 16x(4x^2 + 7)(3x^3 + 1)^4 + (36x^2)(4x^2+7)^2 * (3x^3 + 1)^3
=> y' = [4x(4x^2 + 7)(3x^3 + 1)^3]* [4(3x^3 + 1) + 9x(4x^2+7)]

y= (2x)^(1/2) + (2/x)^(1/2) .......we need the chain rule for each of these
=> y' = (1/2)(2x)^(-1/2) * 2 + (1/2)(2/x)^(-1/2) * (-2x^-2)
=> y' = (2x)^(-1/2) - (x^-2)(2/x)^(-1/2)

y= (x^2-5)^(1/2) * (x^2+3)^(1/3)
.......product rule combined with chain rule again

=> y' = (1/2)(x^2 - 5)^(-1/2) * (2x)(x^2+3)^(1/3) + (x^2-5)^(1/2) * (1/3)(x^2 + 3)^(-2/3) * (2x)
=> y' = x(x^2 - 5)^(-1/2) * (x^2+3)^(1/3) + (2/3)x(x^2-5)^(1/2) * (x^2 + 3)^(-2/3)
• Mar 13th 2007, 02:52 PM
coolaid2u
Thank You!!!! :)
• Mar 13th 2007, 02:56 PM
Jhevon
Quote:

Originally Posted by coolaid2u
Thank You!!!! :)

did you get the steps? do you know how to apply the product and chain rules?
• Mar 13th 2007, 03:08 PM
coolaid2u
Yes, I just wrote the rules wrong :). Thanks!