Differentiate with respect to x

**HNCMATHS** · January 29th 2011, 09:33 AM

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

**e^(i*pi)** · January 29th 2011, 09:35 AM

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 $g(x) = x^3+5$

**FernandoRevilla** · January 29th 2011, 09:37 AM

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

Fernando Revilla

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

**HallsofIvy** · January 29th 2011, 10:52 AM

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

**HNCMATHS** · January 29th 2011, 11:02 AM

Is this the final answer?

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

**HNCMATHS** · January 29th 2011, 11:02 AM

How do I expand on this?

**e^(i*pi)** · January 29th 2011, 11:04 AM

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

**Archie Meade** · January 29th 2011, 11:38 AM

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!

**HNCMATHS** · January 29th 2011, 02:11 PM

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

**mr fantastic** · January 29th 2011, 07:38 PM

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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