# How do you differentiate f(x) = x^2 sqrt((x+1)/(x^2+1))?

Apr 23, 2016

$f ' \left(x\right) = \left({x}^{2}\right) \left(\frac{{\left({x}^{2} + 1\right)}^{\frac{1}{2}} \left(\frac{1}{2 \sqrt{x + 1}}\right) - {\left(x + 1\right)}^{\frac{1}{2}} \left(\frac{x}{\sqrt{{x}^{2} + 1}}\right)}{{x}^{2} + 1}\right) + \left(2 x\right) \left(\sqrt{\frac{x + 1}{{x}^{2} + 1}}\right)$

#### Explanation:

For this problem, we will need a combination of chain rule, product rule, quotient rule, and power rule...Yuck.

$f ' \left(x\right) = \left({x}^{2}\right) \left(\sqrt{\frac{x + 1}{{x}^{2} + 1}}\right) ' + \left(2 x\right) \left(\sqrt{\frac{x + 1}{{x}^{2} + 1}}\right)$

Notice the little apostrophe tagging along the first sqrt((x+1)/(x^2+1)? That means we still need to find the derivative of that part. Start off with chain rule...

$\frac{1}{2 \sqrt{\frac{x + 1}{{x}^{2} + 1}}} \cdot \left(\frac{{\left(x + 1\right)}^{\frac{1}{2}}}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right)\right)$

The first part of the above looks good. Now let's focus on the last part. Using quotient rule...

$\frac{{\left({x}^{2} + 1\right)}^{\frac{1}{2}} \left(\frac{1}{2 \sqrt{x + 1}}\right) - {\left(x + 1\right)}^{\frac{1}{2}} \left(\frac{x}{\sqrt{{x}^{2} + 1}}\right)}{{x}^{2} + 1}$

Dang...I would like to see anyone simplify this...your answer , by the way, is

$f ' \left(x\right) = \left({x}^{2}\right) \left(\frac{{\left({x}^{2} + 1\right)}^{\frac{1}{2}} \left(\frac{1}{2 \sqrt{x + 1}}\right) - {\left(x + 1\right)}^{\frac{1}{2}} \left(\frac{x}{\sqrt{{x}^{2} + 1}}\right)}{{x}^{2} + 1}\right) + \left(2 x\right) \left(\sqrt{\frac{x + 1}{{x}^{2} + 1}}\right)$

Many apologies that I can't go any further into this problem. This is certainly a mammoth...just hope that something like this doesn't appear on your tests. :)