differentiation using quotient rule - basic equation

Question

Having trouble understanding the end result of this, so if someone could explain it to me that would be great :).

So I have the function $f(x)=\frac{\sqrt{x}}{\sin(x)}$

My steps after taking the derivative of $\frac{\sqrt{x}}{\sin(x)}$:

$$\frac{\sin(x)-\cos(x)\sqrt {x}}{2\sqrt x*\sin^2(x)}$$

Now I do not understand what I did wrong here, for some reason the correct result is $$\frac{\sin(x)-2x*\cos(x)}{2\sqrt{x}*\sin^2(x)}$$

Would appreciate any input.

Answer 1

Using the formula $$\left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}$$ we get \begin{align*} \left(\frac{\sqrt{x}}{\sin(x)}\right)' &=\frac{\sin(x)\left(\frac{1}{2\sqrt{x}}\right)-\sqrt{x}\cos(x)}{\sin^2(x)}\\[4pt] &=\frac{\sin(x)\left(\frac{1}{2\sqrt{x}}\right)-\sqrt{x}\cos(x)}{\sin^2(x)} \cdot \frac{2\sqrt{x}}{2\sqrt{x}} \\[4pt] &=\frac{\sin(x)-2x\cos(x)}{2\sqrt{x}\sin^2(x)} \\[4pt] \end{align*}

Answer 2

Look at the first thing you did. The derivative of $\sqrt{x}$ is certainly not equal to 1. Also, the derivative of $sin(x)$ is $cos(x)$, not $-cos(x)$.

Answer 3

Recall the quotient rule:

$$\frac{\mathrm d}{\mathrm d x}\left(\frac{f(x)}{g(x)}\right) = \frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{[g(x)]^2}. $$

Here, $f(x)=\sqrt x$ and $g(x) = \sin x$. We also have that

\begin{align*} f^\prime(x) &= \frac{1}{2\sqrt x}, \\ g^\prime(x) &= \cos x. \end{align*}

You should be able to do the rest from here.

Answer 4

You tried to do two steps at once in the first step and messed it up:

$$\frac{\frac{\sin x}{2\sqrt{x}} - \cos x \cdot \sqrt{x}}{\sin^2 x} \neq \frac{1\cdot\sin x - \cos x \cdot \sqrt{x}}{2\sqrt{x}\sin^2 x}$$