# Math Help - Determing the derivative of a function using the chain rule

1. ## Determing the derivative of a function using the chain rule

How do I get from the problem to the answer using the chain rule?
It would be much appreciated if someone indicated the steps for me, thank you!

2. Originally Posted by !!!
How do I get from the problem to the answer using the chain rule?
It would be much appreciated if someone indicated the steps for me, thank you!
To differentiate the product you use the product rule. As part of that process you need to differentiate $(x^2 + 2)^{1/3}$. Get that derivative using the chain rule.

3. Only one comment to add: you recently posted the same problem on MathLinks. They don't actually reply as faster as we can, so I suggest you to stay here 'cause you'll get a quickly answer and full solutions. (This last thing depends of your problem.)

4. Alright, thanks for the help and heads-up, guys.
I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:

How do I get from that to the correct answer of
?

5. Originally Posted by !!!
Alright, thanks for the help and heads-up, guys.
I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:

How do I get from that to the correct answer of
?
Note that $(x^2 + 2)^{-2/3} = (x^2 + 2)^{1/3} (x^2 + 2)^{-1} = \frac{(x^2 + 2)^{1/3}}{(x^2 + 2)}$. Make that substitution and then factorise by taking out the common factor of $2x (x^2 + 2)^{1/3}$ in each term.

6. Originally Posted by !!!
Alright, thanks for the help and heads-up, guys.
I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:

How do I get from that to the correct answer of
?
$y'=2x(x^2+2)^\frac{1}{3}+(x^2+1)\cdot\frac{1}{3}(x ^2+2)^\frac{-2}{3}(2x)$

rewriting the second term with a positive exponent gives

$y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\cdot\frac{1}{ 3(x^2+2)^\frac{2}{3}}$

multiplying the numerator and denominator of the second term

[tex]MATH]

simplifying

$y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\frac{(x^2+2)^ \frac{1}{3}}{3(x^2+2)}$

factoring out the GCF gives

$y'=2x(x^2+2)^\frac{1}{3}\left[1+\frac{x^2+1}{3(x^2+2)}\right]$

7. Thank you very much, everyone. Now I understand how to do that problem.
I also have trouble determining the derivatives of these and I'm not sure if I'm doing them correctly because the book don't have their answers.

$1. y = x sin x^{1/2}$
$y' = x cos (x^{1/2}) d/dx (x^{1/2}) + (sin x^{1/2}) d/dx (x^{1/2})$
$y' = x cos \sqrt{x} + sin \sqrt{x} / {2x \sqrt{x}}$

2. $y = x/(7-3x^{1/2})$
I only got up to $(7-3x)^{1/2} (-3) - x[(1/2)(7-3x^{1/2})] (-3)$

8. Originally Posted by !!!
Thank you very much, everyone. Now I understand how to do that problem.
I also have trouble determining the derivatives of these and I'm not sure if I'm doing them correctly because the book don't have their answers.

$1. y = x sin x^{1/2}$
$y' = x cos (x^{1/2}) d/dx (x^{1/2}) + (sin x^{1/2}) d/dx (x^{1/2})$
$y' = x cos \sqrt{x} + sin \sqrt{x} / {2x \sqrt{x}}$

2. $y = x/(7-3x^{1/2})$
I only got up to $(7-3x)^{1/2} (-3) - x[(1/2)(7-3x^{1/2})] (-3)$
1. You need the product rule:

$u = x \Rightarrow \frac{du}{dx} = 1$

$v = \sin x^{1/2} \Rightarrow \frac{dv}{dx} = \cos x^{1/2} \times \frac{1}{2} x^{-1/2} = \cos \sqrt{x} \times \frac{1}{2\sqrt{x}}$, where the chain rule has also been used.

Then you substitute the above results into $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.

Some simplification of the resulting answer will be possible .....

----------------------------------------------------------------------------------------------

2. You need the quotient rule:

$u = x \Rightarrow \frac{du}{dx} = 1$

$v = 7 - 3 x^{1/2} \Rightarrow \frac{dv}{dx} = -\frac{3}{2} x^{-1/2} = -\frac{3}{2\sqrt{x}}$.

Then you substitute the above results into $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.

Some simplification (requiring a little bit of algebra) of the resulting answer will be desirable .....