find y' when y=x^2\sqrt{5x-1}

[havenoclu]

Find y' when:

\(\displaystyle \L y=x^2\sqrt{5x-1}\)

I don't know if my answer is correct. The square root is throwing me. I'm not sure how to get properly put it into this derivative.

I attempted to "simplify" the square root and perform the chain rule on it. Here's my answer:

\(\displaystyle \L y'=(2x)(2\frac{1}{2}(5x-1)^{-1/2})\)

 

[galactus]

The product rule:

\(\displaystyle \L\\x^{2}(\frac{1}{2})(5x-1)^{\frac{-1}{2}}(5)+(5x-1)^{\frac{1}{2}}(2x)\)

=\(\displaystyle \L\\2x\sqrt{5x-1}+\frac{5x^{2}}{2\sqrt{5x-1}}\)

 

[havenoclu]

ok..great...appreciate it. do you think it's necessary to put it back into square root form? I'm horrible at that. I see how you came up with the first equation (using product/chain rules), but as far as "simplifying" that equation with the square roots I'm lost.

 

[tkhunny]

How can yuo be horrible at it? That makes absolutely no sense.

Use whichever form 1) makes better sense to you, or 2) you are required to use. Familiarity with both is essential.