# Why is my answer incorrect for this differentiation question?

$$y = x* ((x^2+1)^{1/2})$$ I must find $$dy/dx$$

$$u = x, v = (x^2+1)^{1/2}$$ To do this I must use the product rule and the chain rule. To get dv/dx, $$(dv/dx) = (1/2)*(b)^{-1/2}*2x$$ $$(dv/dx) = x*(b)^{-1/2}$$ $$(dv/dx) = x*(x^2+1)^{-1/2}$$ $$(dv/dx) = x*\frac{1}{\sqrt{x^2+1}}$$ $$(dv/dx) = \frac{x}{\sqrt{x^2+1}}$$ so now $$u*(dv/dx) = x*\frac{x}{\sqrt{x^2+1}}$$ $$u*(dv/dx) = \frac{x^2}{\sqrt{x^2+1}}$$ now for v* du/dx

$$v*du/dx = 1 * ((x^2+1)^{1/2}) =(x^2+1)^{1/2}$$

so adding the parts together as follows : $$(u*(dv/dx))+(v*(du/dx))$$

gives: $$\frac{x^2}{\sqrt{x^2+1}} +(x^2+1)^{1/2}$$

which could be shown as

$$\frac{x^2}{\sqrt{x^2+1}} + \frac{(x^2+1)^{1/2}}{1}$$

so far I feel confident with my workings out, here is what I do next.

$$\frac{(2x^4 +x^2)^{1/2}}{\sqrt{x^2+1}}$$

as I have multiplied the numerators together. This answer is incorrect as I am aware the correct answer is : $$\frac{2x^2+1}{(x^2+1)^{1/2}}$$

Can someone please show me where I have gone wrong? Also show the correct course of action to solve this ? I realise the denominator is the same. Any help is very much appreciated. Thanks

• When you first compute $dv/dx$, you set $b = x^{2} + 1$. Then, when you substituted back, you put $b = x^{2} - 1$. May 7, 2016 at 14:45
• @Mattos , My apologies it was meant to be +1 throughout. I have corrected this. May 7, 2016 at 15:05
• You complicate it too much that it's even difficult for someone to correct it. Check down below for an easy way to computing the derivative. May 7, 2016 at 15:07
• @CharalamposFilippatos I will try to, I want to make it clear for the reader to see where I have gone wrong, also I am not skilled in this topic so it is difficult for me to miss steps. May 7, 2016 at 15:30
• Check my solution down below mate ! Don't forget to vote up and approve if it makes you understand what you asked. The way I solve it is by different notation but it's the chain rule that is applied. Check it out. May 7, 2016 at 15:52

As Mattos said, the problem uses $x^{2} + 1$ and you used $x^{2} - 1$. Otherwise you were ok to the point where you said you were confident. At that point you have two fractions to add. You need to get a common denominator and add the numerators, not multiply the numerators.
• To get an LCD you need to factor all denominators first. Then the LCD is the product of each distinct factor. In your case, the two denominators are factored, and one of them is 1; so your LCD is simply $\sqrt{x^2+1}$. Just multiply the 2nd fraction by $$\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}$$ That will simplify your numerator. May 11, 2016 at 6:32
So if $f(x) = x(x^2 +1)^{\frac{1}{2}}$
Then : $f'(x) = [x(x^2 +1)^{\frac{1}{2}}]' = x'(x^2 +1)^{\frac{1}{2}} + x[(x^2 +1)^{\frac{1}{2}}]' = (x^2 +1)^{\frac{1}{2}} + x[\sqrt{x^2+1}]' = (x^2 +1)^{\frac{1}{2}} + x \frac{1}{2\sqrt{x^2+1}}[x^2+1]' = (x^2 +1)^{\frac{1}{2}} + x \frac{1}{2\sqrt{x^2+1}}2x = (x^2 +1)^{\frac{1}{2}} + 2x^2 \frac{1}{2\sqrt{x^2+1}}$ If you fix the fractions, you get the answer you want.