find y' for y=3x/sqrt(x-4): check if correct please

[crzymath]

problem: find first derivative: y=3x/sqrt(x-4)

y'=(x-4)[sup:2nry7soz]1/2[/sup:2nry7soz](3)-3x(1/2x-4)[sup:2nry7soz]-1/2[/sup:2nry7soz]/(x-4) <---- Quotient Rule
y'=sqrt(3x-12)-((3x[sup:2nry7soz]2[/sup:2nry7soz]/2)-12x)[sup:2nry7soz]-1/2[/sup:2nry7soz]/(x-4)
y'=sqrt(3x-12)/sqrt(((3x[sup:2nry7soz]2[/sup:2nry7soz]/2)-12x))(x-4)

 

[mmm4444bot]

No, it's not correct.

It looks like you need to brush up on your algebra skills.

 

[crzymath]

Re: check if this is correct please

can u at least tell me what the answer should be so i can go back and redo the problem?

 

[crzymath]

Re: check if this is correct please

if not the answer-can u tell me what i did wrong? clearly its an algebra error-but where??

 

[galactus]

If possible, use the product rule instead of the quotient rule.

\(\displaystyle \frac{3x}{\sqrt{x-4}}=\overbrace{3x}^{\text{f(x)}}\cdot\underbrace{(x-4)^{\frac{-1}{2}}}_{\text{g(x)}}\)

Product rule:

\(\displaystyle \overbrace{3x}^{\text{f(x)}}\cdot\underbrace{(\frac{-1}{2})(x-4)^{\frac{-3}{2}}}_{\text{g'(x)}}+\overbrace{(x-4)^{\frac{-1}{2}}}^{\text{g(x)}}\cdot\underbrace{(3)}_{\text{f'(x)}}\)

No need to worry about the chain rule because the derivative of x-4 is just 1.

Factor out \(\displaystyle \frac{3}{2}(x-4)^{\frac{-3}{2}}\)

\(\displaystyle y'=\frac{3}{2}(x-4)^{\frac{-3}{2}}\left[-x+2(x-4)\right]\)

\(\displaystyle y'=\frac{3(x-8)}{2(x-4)^{\frac{3}{2}}}\)