Why is my answer incorrect for this differentiation question?

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

Answer 1

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.

Answer 2

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.