Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
LaTeX tip: don't use period for multiplication. Use \cdot. –  kahen Oct 25 '12 at 14:58
add comment

5 Answers 5

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:

$$\left[x^2\sqrt{2x+5}\right]'=x^2\left[\sqrt{2x+5}\right]'+\left[x^2\right]'\sqrt{2x+5}\;.$$

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.

share|improve this answer
    
$2x.\sqrt{2x+5} + x^2 . \dfrac{1}{2\sqrt{2x+5}}$ ? –  ZafarS 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.. –  ZafarS 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: $2x.\sqrt{2x+5} + x^2 . \dfrac{2}{2\sqrt{2x+5}}$ –  ZafarS Oct 25 '12 at 14:58
show 3 more comments

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

share|improve this answer
    
I spotted that you used eqnarray. Don't do that. –  kahen Oct 25 '12 at 14:57
add comment

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} $

share|improve this answer
    
I had $2x.\sqrt{2x+5} + x^2 . \dfrac{1}{2\sqrt{2x+5}}$ –  ZafarS 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
show 2 more comments

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

share|improve this answer
add comment

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} $

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.