Derivative: Quotient Rule, Numerator is the square root of X, Can't get right answer

[Skyphon Pox]

I'm having a problem getting my textbook's answer for this question. Can you please help?

Find the derivative of x^(1/2) / (x^2 + 3)

Every time I do it, my answer is the same : (x + 3 - 4x^2) / [2x^(1/2) ] [x^2 + 3]^2

But my textbook says the answer is (3 - x^2) / [x^(1/2)] [x^2+3]^2

Can you please show me step by step how to get to this solution? It seems to me like I'm doing absolutely everything right, so I have no idea what I'm doing wrong, if anything.

 

[Deleted member 4993]

I'm having a problem getting my textbook's answer for this question. Can you please help?

Find the derivative of x^(1/2) / (x^2 + 3)

Every time I do it, my answer is the same : (x + 3 - 4x^2) / [2x^(1/2) ] [x^2 + 3]^2

But my textbook says the answer is (3 - x^2) / [x^(1/2)] [x^2+3]^2

Can you please show me step by step how to get to this solution? It seems to me like I'm doing absolutely everything right, so I have no idea what I'm doing wrong, if anything.

we know:

\(\displaystyle \dfrac{d}{dx}\left [\dfrac{u(x)}{v(x)}\right ] \ = \ \dfrac{u'*v - v'*u}{v^2}\)

u = x^1/2 → u' = (1/2)x^-1/2

v = x² + 3 → v' = 2x

\(\displaystyle \dfrac{d}{dx}\left [\dfrac{u(x)}{v(x)}\right ] \ = \ \dfrac{u'*v - v'*u}{v^2} \)

\(\displaystyle = \ \dfrac{[(1/2)x^{-1/2}]*(x^2 + 3) - 2x*x^{1/2}}{(x^2+3)^2}\)

now continue...

 

[Skyphon Pox]

First, THANK YOU FOR YOUR INTEREST.

Second, that's exactly what I got.

My problem lies in getting from the exact last equation you stated down to the textbook answer I provided. Can you show me how that might happen?

Like I say, I feel I solved the problem using all the right rules and techniques, but my final answer does not match the textbook.

 

[mmm4444bot]

my textbook says the answer is (3 - x^2) / ([x^(1/2)] [x^2+3]^2)

This textbook answer is not correct.


My numerator works out to:

\(\displaystyle -\frac{3}{2} \, (x^2 - 1)\)


which expands to:

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


There is no way to factor this result such that only 3 - x^2 appears in the numerator of the answer.

Every time I do it, my answer is the same : (x + 3 - 4x^2) / [2x^(1/2) ] [x^2 + 3]^2

This expression does not look correct, to me, either.

You wrote that the last expression in Subhotosh's reply matches your result exactly. If this is the case, then you've got the correct derivative.

If you're trying to simplify that expression, please show your steps so that we might see what you're doing wrongly.

 

[Skyphon Pox]

First,

(X^1/2)’(x²+3)-(x^1/2)(x²+3)’ / (x²+3)²

 

[Skyphon Pox]

=(½)x^-1/2(x²+3) – x^1/2(2x) / (x²+3)²


Then I multiply everything by two to get rid of the fraction in the numerator:


=2(½)x^-1/2(x²+3) – 2(x^1/2)(2x) / 2(x²+3)²

 

[Skyphon Pox]

=x^-1/2(x²+3) – x^1/2(4x) / 2(x²+3)²

Then I factor out x^-1/2:

= x^-1/2 [[ [ x² + 3] – x(4x) ]] / 2(x²+3)²

 

[Skyphon Pox]

Then I get

=x^-1/2(x² + 3 – 4x²) / 2(x²+3)²


=( 3 – 3x² )/ 2(x^1/2)(x²+3)²


Okay, so I found an error, AND this new response is in line with the answer they gave me in another forum. But I'm still troubled that my textbook doesn't match this answer.

 

[lookagain]

my answer is the same : (x + 3 - 4x^2)/{[2x^(1/2)][x^2 + 3]^2}but my textbook says the answer is (3 - x^2)/{[x^(1/2)][x^2 + 3]^2}

first, [(x^1/2)’(x² + 3) - (x^1/2)(x² + 3)’ ]/(x² + 3)²

then i get

= x^-1/2(x² + 3 – 4x²)/2(x² + 3)²


= ( 3 – 3x² )/2(x^1/2)(x² + 3)²

<sup>\(\displaystyle \text{skyphon pox, a couple of these show the needed grouping symbols.}\)
\(\displaystyle \text{The others, one shown, but another post is not, could be fixed, too.}\)

Because of the Order of Operations, you must have them.</sup>

 

[mmm4444bot]

(3 – 3x²)/[2(x^1/2)(x²+3)²]

I'm still troubled that my textbook doesn't match this answer.

I'm slightly-troubled that you're still not typing grouping symbols around the denominator. Did you notice the earlier corrections?

I already told you that the text's answer in your original post is wrong.

Math texts are riddled with errors. Try not to let this fact bother you. Find satisfaction in proving the text wrong, instead. :cool:

 

[lookagain]

I already told you that the text's answer in your original post is wrong.

Math texts are riddled with errors. Try not to let this fact bother you. Find satisfaction in proving the text wrong, instead. :cool:

Also, the 5th, 6th, and 7th posts on this thread indicate that the OP doesn't
know that those certain numerators also need grouping symbols around them.

 

[camreron tech]

I'm having a problem getting my textbook's answer for this question. Can you please help?

Find the derivative of x^(1/2) / (x^2 + 3)

Every time I do it, my answer is the same : (x + 3 - 4x^2) / [2x^(1/2) ] [x^2 + 3]^2

But my textbook says the answer is (3 - x^2) / [x^(1/2)] [x^2+3]^2

Can you please show me step by step how to get to this solution? It seems to me like I'm doing absolutely everything right, so I have no idea what I'm doing wrong, if anything.

well what you should see it you can rest youinternet how
click off wait sec and then turn computer off then turn back on in sec

 

[Deleted member 4993]

Originally Posted by Skyphon Pox
I'm having a problem getting my textbook's answer for this question. Can you please help?

Find the derivative of x^(1/2) / (x^2 + 3)

Every time I do it, my answer is the same : (x + 3 - 4x^2) / [2x^(1/2) ] [x^2 + 3]^2

But my textbook says the answer is (3 - x^2) / [x^(1/2)] [x^2+3]^2

Can you please show me step by step how to get to this solution? It seems to me like I'm doing absolutely everything right, so I have no idea what I'm doing wrong, if anything.

we know:

\(\displaystyle \dfrac{d}{dx}\left [\dfrac{u(x)}{v(x)}\right ] \ = \ \dfrac{u'*v - v'*u}{v^2}\)

u = x^1/2 → u' = (1/2)x^-1/2

v = x² + 3 → v' = 2x

\(\displaystyle \dfrac{d}{dx}\left [\dfrac{u(x)}{v(x)}\right ] \ = \ \dfrac{u'*v - v'*u}{v^2} \)

\(\displaystyle = \ \dfrac{[(1/2)x^{-1/2}]*(x^2 + 3) - 2x*x^{1/2}}{(x^2+3)^2}\)

\(\displaystyle = \ \dfrac{(x^2 + 3) - 4x^2 }{2*(x^2+3)^2*x^{1/2}}\)

\(\displaystyle = \ \dfrac{3(1 - x^2)}{2*(x^2+3)^2*x^{1/2}}\)