Compute the derivative

Question

Compute the derivatives of: $$ y = {\sqrt{x + \sqrt{x}}} $$

My attempt:

$$=\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+\sqrt{x}\right)$$

$$\frac{d}{du}\left(\sqrt{u}\right)$$

$$=\frac{d}{du}\left(u^{\frac{1}{2}}\right)$$

$$=\frac{1}{2}u^{\frac{1}{2}-1}$$

$$ \frac{1}{2}u^{-\frac{1}{2}} $$

$$ =u^{-\frac{1}{2}} $$

$$ =\frac{1}{2}u^{-\frac{1}{2}}$$

$$=\frac{1}{2}\cdot \frac{1}{\sqrt{u}}$$

$$=\frac{1\cdot \:1}{2\sqrt{u}}$$

$$=\frac{1}{2\sqrt{x}}$$

$$=1+\frac{1}{2\sqrt{x}}$$

$$=\frac{1}{2\sqrt{u}}\left(1+\frac{1}{2\sqrt{x}}\right)$$

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

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

$$=\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}$$

$$1+\frac{1}{2\sqrt{x}}$$

$$=\frac{1}{1}+\frac{1}{2\sqrt{x}}$$

$$\frac{1}{1}+\frac{1}{2\sqrt{x}}$$

$$=\frac{1\cdot \:2\sqrt{x}}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}$$

$$=\frac{1\cdot \:2\sqrt{x}+1}{2\sqrt{x}}$$

$$=2\sqrt{x}+1$$

$$=\frac{2\sqrt{x}+1}{2\sqrt{x}}$$

$$=\frac{\frac{2\sqrt{x}+1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}$$

$$=\frac{2\sqrt{x}+1}{2\cdot \:2\sqrt{x}\sqrt{x+\sqrt{x}}}$$

$$=\frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}}$$

$$\frac{d}{dx}\left(y\right)=\frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}}$$

I did not get full marks on this question, and I would like to know where I have gone wrong.
(Please note that I cannot get feedback from my professor as not every student has submitted their assignment). Thank you.

Answer 1

You need to use the chain rule, and properly identify the components.

You have $f(x)=g\Big(h(x)\Big)$, where $g(x)=\sqrt{x}$ and $h(x)=x+\sqrt{x}$.

Then $f'(x)=g'\Big(h(x)\Big)h'(x)$.

The end result is thus $f'(x)=\frac{1 + \frac{1}{2 \sqrt{x}}}{2 \sqrt{\sqrt{x} + x}}$.