# derivative with square root

I have been trying to figure this equation for some time now, but have come up empty. I have tried multiple ways on solving it. Whether by using the Quotient Rule or some other method, I can't seem to figure it out. Any help would be appreciated.

Find the derivative of the function

$y = \frac{x^{2} + 8x + 3 }{\sqrt{x}}$

$y' = \frac{3x^{2}+ 8x - 3 }{2x^\frac{3}{2}}$

I'm having trouble getting to that answer. So if someone could point me in the right direction by showing the steps, I would be grateful. Thanks!

• If you show what you are doing, it will make it easier for someone here to see why you are having trouble... – colormegone Jun 5 '16 at 0:30
• I used the quotient rule, but I end up at a dead end, or maybe it's staring me in the face. – Carter Hunt Jun 5 '16 at 0:31
• Be careful with factoring, with exponents/roots, and particularly with distributing minus-signs. (And, yes, I was wondering where the factor of 2 in the denominator had gone...) – colormegone Jun 5 '16 at 0:32
• Yea, I'm having trouble with the whole square root symbol in the bottom. – Carter Hunt Jun 5 '16 at 0:37
• One of your numerator terms in the derivative is $\ \frac{3}{2 \ \sqrt{x}} \$ . The $\ \frac{1}{2 \ \sqrt{x}} \$ is being factored out of all the terms in the numerator so it can be "moved" to the denominator, where it joins the $\ x \$ already there. – colormegone Jun 5 '16 at 0:40

$y = \frac{x^{2} + 8x + 3 }{\sqrt{x}}$
You are right in the direction of using quotient rule. $$y'=\frac{(2x+8)\sqrt{x}-(x^2+8x+3)\frac{1}{2\sqrt{x}}}{(\sqrt x)^2}$$Multiply by $\frac{2\sqrt x}{2\sqrt x}$ to get rid of the $\frac{1}{2\sqrt{x}}$ $$y'=\frac{4x^2+16x-x^2-8x-3}{2x\sqrt x}$$And $2x\sqrt x=2x^{\frac32}$, So $$y'=\frac{3x^{2}+ 8x - 3 }{2x^\frac{3}{2}}$$

While using the quotient rule for $f(x)=\dfrac{h(x)}{g(x)}\rightarrow f'(x)=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}$ is straightforward and gives

$y'=\dfrac{(2x+8)\sqrt{x}-\frac{1}{2}(x^2+8x+3)x^{-\frac{1}{2}}}{x}=(2x+8)x^{-\frac{1}{2}}-\frac{1}{2}(x^2+8x+3)x^{-\frac{3}{2}}=x^{-\frac{3}{2}}\left[(2x+8)x-\frac{1}{2}(x^2+8x+3)\right]=\dfrac{3x^2+8x-3}{2x^{-\frac{3}{2}}}$

we can also, with a simple manipulation, avoid the quotient rule:

$y=x^{\frac{3}{2}}+8\sqrt{x}+3x^{-\frac{1}{2}}$

$y'=\frac{3}{2}x^{\frac{1}{2}}+4x^{-\frac{1}{2}}-\frac{3}{2}x^{-\frac{3}{2}}=\frac{3x^2+8x-3}{2x^{-\frac{3}{2}}}$

For differentiating $\sqrt x$ one should rewrite it as $x^{\frac{1}{2}}$ and use the power rule. Given $y=\frac{x^2+8x+3}{\sqrt x}$ we have: $$y'=\frac{\left(\sqrt x\right)\left(2x+8\right)-\left(\frac{1}{2}x^{-\frac{1}{2}}\right)\left(x^2+8x+3\right)}{\left(\sqrt x\right)^2} \\=x^{-\frac{3}{2}}\left(2x^2+8x-\frac{1}{2}x^2-4x-\frac{3}{2}\right) \\=\frac{1}{2}x^{-\frac{3}{2}}\left(3x^2+8x-3\right)$$

Instead of using the quotient rule, you could use the fact that $\sqrt{x}=x^{1/2}$.

Then we get $y$ in the form of a polynomial:

$$y = \frac{x^{2} + 8x + 3 }{\sqrt{x}}=x^{-1/2}(x^2+8x+3)=x^{3/2}+8x^{1/2}+3x^{-1/2}$$

Taking the derivative then gives. $$y' = \frac{3}{2}x^{1/2}+\frac{1}{2}\cdot 8x^{-1/2}-\frac{1}{2}\cdot 3x^{-3/2} = \frac{3x^{4/2}}{2x^{3/2}}+\frac{8x}{2x^{3/2}} -\frac{3}{x^{3/2}} = \frac{3x^{2}+ 8x - 3 }{2x^\frac{3}{2}}$$

There is another way which makes life easier when you face product and/or quotients : it is logarithmic differentiation.

Consider $$y = \frac{x^{2} + 8x + 3 }{\sqrt{x}}$$ Take logarithms $$\log(y)=\log(x^{2} + 8x + 3)-\frac 12 \log(x)$$ Differentiate $$\frac {y'}y=\frac{2x+8}{x^{2} + 8x + 3 }-\frac 1{2x}=\frac{3x^2+8x-3}{2x(x^{2} + 8x + 3 )}$$ $$y'=\frac{3x^2+8x-3}{2x(x^{2} + 8x + 3 )} \times\frac{x^{2} + 8x + 3 }{\sqrt{x}}=\frac{3x^2+8x-3}{2x \sqrt x}$$