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How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$

I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$

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LaTeX tip: don't use period for multiplication. Use \cdot. – kahen Oct 25 '12 at 14:58
up vote 2 down vote accepted

Work from the outside in. Begin by differentiating it term by term: $$f\,'(x)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-\frac{d}{dx}(6)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-0\;.$$

Now you have to calculate the derivative of $x^2\sqrt{2x+5}$. This is a product, so you use the product rule:


To complete the differentiation you’ll need the derivative of $x^2$, which is very easy, and the derivative of $\sqrt{2x+5}$. That one is also pretty easy once you rewrite the function as $(2x+5)^{1/2}$: the power rule and the chain rule will take care of it.

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$2x.\sqrt{2x+5} + x^2 . \dfrac{1}{2\sqrt{2x+5}}$ ? – JohnPhteven Oct 25 '12 at 14:53
@ZafarS: Not quite: the derivative of $(2x+5)^{1/2}$ is $\frac12(2x+5)^{-1/2}[2x+5]'$, and you forgot that last factor; it comes from the chain rule. – Brian M. Scott Oct 25 '12 at 14:55
So that's where the 2 comes from.. – JohnPhteven Oct 25 '12 at 14:56
@ZafarS: Yes, exactly. And if it had been $2x^2+5$ inside the square root, for instance, you’d have got an extra factor of $4x$ from the chain rule. – Brian M. Scott Oct 25 '12 at 14:57
So is it: $2x.\sqrt{2x+5} + x^2 . \dfrac{2}{2\sqrt{2x+5}}$ – JohnPhteven Oct 25 '12 at 14:58

$$ \begin{eqnarray*} y &=& x^2(2x+5)^{1/2} - 6 \\ \frac{dy}{dx} &=& \frac{d}{dx} \Big[ x^2(2x+5)^{1/2} \Big] - \frac{d}{dx}\Big[ 6 \Big], \qquad \textrm{Sum/Difference Rule}\\ &=& \frac{d}{dx}\Big[x^2\Big](2x+5)^{1/2} + x^2\frac{d}{dx}\Big[(2x+5)^{1/2}\Big] - 0, \qquad \textrm{Product Rule}\\ &=& 2x(2x+5)^{1/2} + x^2\left( \frac{1}{2}(2x+5)^{-1/2}\cdot 2\right), \qquad \textrm{Chain Rule}\\ &=& 2x\sqrt{2x+5} + \frac{ x^2}{\sqrt{2x+5}}, \qquad \textrm{simplification.} \end{eqnarray*} $$

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I spotted that you used eqnarray. Don't do that. – kahen Oct 25 '12 at 14:57

Using the chain rule where $\cfrac {df}{dx} = \cfrac {dg}{du} \cfrac {du}{dx} $ if $f(x) = g(u(x))$ and the product rule where $\cfrac {d}{dx} (uv) = v\cfrac {du}{dx} + u\cfrac {dv}{dx} $

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I had $2x.\sqrt{2x+5} + x^2 . \dfrac{1}{2\sqrt{2x+5}}$ – JohnPhteven Oct 25 '12 at 14:48
Sounds like you need the product rule instead. – gt6989b Oct 25 '12 at 14:50
redo it carefully – user31280 Oct 25 '12 at 14:52
@ZafarS: What you have is almost right, but you forgot the chain rule when you differentiated $(2x+5)^{1/2}$: you should get $\frac12(2x+5)^{-1/2}\cdot[2x+5]'$, and you’re missing that last factor. – Brian M. Scott Oct 25 '12 at 14:53
@BrianM.Scott that was probably for ZafarS, not for me :) – gt6989b Oct 25 '12 at 14:56

$2x . \sqrt{2x+5} + \dfrac{x^2}{2} . \dfrac{1}{\sqrt{2x+5}}2$

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I would use the product rule, which you seem to have tried from your comment above. Here would be the idea:

$\begin{split} \frac{d[x^2 \sqrt{2x+5} - 6]}{dx} &= \frac{dx^2}{dx} \sqrt{2x+5} + x^2 \frac{d(2x+5)^{\1/2}}{dx} \\ &= 2x \sqrt{2x+5} + x^2 \frac{1}{2} (2x+5)^{-1/2} \cdot 2 \\ &= 2x \sqrt{2x+5} + \frac{x^2}{\sqrt{2x+5}} \end{split} $

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