Thread: Derivative help: f(x) = 6x^6 sqrt[x] + 2/(x^2 sqrt[x])

I know how to take the Derivative for a much simpler question but I have no idea how to do it with this one. Do I get rid of the square root first or something?
This is what I did:







and I came up with the answer of:



but its wrong, can someone help?

Originally Posted by killasnake

I know how to take the Derivative for a much simpler question but I have no idea how to do it with this one. Do I get rid of the square root first or something? ...

Hi,

I've attached an image to show you what I've calculated.

EB

Thank you so much for the help!

Originally Posted by killasnake

I know how to take the Derivative for a much simpler question but I have no idea how to do it with this one. Do I get rid of the square root first or something?
This is what I did:







and I came up with the answer of:



but its wrong, can someone help?

(I assume) Earboth's answer is correct, but doesn't address what you did wrong.

It appears you tried to take the derivative of the first term and didn't do anything with the second. (At least, I'm assuming the second line is supposed to be the derivative. You left it as "f(x)" not "f'(x)".) Also in your first term you didn't use the product rule.

Let me run through the derivative term by term.

Start with the derivative of 6x^6*sqrt(x).

The derivative would be:
6*[6x^5]*sqrt(x) + 6x^6*[(1/2)*1/sqrt(x)] = (36*x^6 + 3x^6)/sqrt(x) = 39x^6/sqrt(x)
(The [ ] indicates where I've taken the derivative of a term.)

The derivative of the second term, 2/{x^2*sqrt(x)}:
Think of this as 2/x^2 * x^(-1/2)
The derivative would be:
2*[(-2)/x^3]*x^(-1/2) + (2/x^2)*[(-1/2)*x^(-3/2)] = -5/{x^3*sqrt(x)}
(Where x^(-3/2) = 1/{x*sqrt(x)}.)

So the derivative of the overall expression would be:
39x^6/sqrt(x) + 5/{x^3*sqrt(x)}
(Which is the same as Earboth's answer.)

Needless to say, Earboth's method is simpler...

-Dan