# Thread:

1. ## Differentiate with respect to x

Y = $\displaystyle \sqrt{(x^3+5)}$

2. Use the chain rule: $\dfrac{d}{dx} f[g(x)] = f'[g(x)] \cdot g'(x)$

In your case you have $f(x) = \sqrt{g(x)}$ and $g(x) = x^3+5$

3. $\dfrac{d}{dx}\left(\sqrt{u(x)}\right)=\dfrac{1}{2\ sqrt{u(x)}}\cdot u'(x)$

Fernando Revilla

Edited: Sorry, I didn't see e^(i*pi)'s post.

4. You can also write this as $(x^3+5)^{1/2}$ to use the "power law" and the chain rule.

5. Is this the final answer?

$=\dfrac{3 x^2}{2\sqrt{(x^3 + 5)}}$

6. How do I expand on this?

7. Yes that's fine, I wouldn't bother expanding that though.

8. Originally Posted by HNCMATHS
Is this the final answer?

$=\dfrac{3 x^2}{2\sqrt{(x^3 + 5)}}$
Let's see!

$\displaystyle\ u=x^3+5\Rightarrow\frac{du}{dx}=3x^2$

From the Chain Rule

$\displaystyle\frac{dy}{dx}=\frac{dy}{du}\;\frac{du }{dx}=\frac{d}{du}u^{\frac{1}{2}}\;3x^2=\frac{1}{2 }u^{-\frac{1}{2}}\;3x^2=\frac{1}{2u^{\frac{1}{2}}}3x^2= \frac{3x^2}{2\sqrt{x^3+5}}$

So you see, those fast methods can be quite useful as an alternative to messing with fractional indices!

9. Originally Posted by HallsofIvy
You can also write this as $(x^3+5)^{1/2}$ to use the "power law" and the chain rule.
HI Halls of Ivy it is this method I would like you to expand on please, Thanks

10. Originally Posted by HNCMATHS
HI Halls of Ivy it is this method I would like you to expand on please, Thanks
Have you been taught the chain rule? Your class notes and textbook will have many examples for you to follow and relate to the question you posted.
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