# Let $f(x)=\frac{2x+1}{\sin(x)}$ find $f'(x)$

Let $$f(x)=\frac{2x+1}{\sin(x)}$$ Find $f'(x).$

I used Quotient Rule
\begin {align*}\frac{\sin(x)2-(2x+1)\cos(x)}{\sin^2(x)}\\ =\frac{3-2x\cos(x)}{\sin(x)} \end {align*}

Is that right? I don't know how to get the answer. Please help me out, thanks.

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Please check what I edit in your question. – DonAntonio Jun 3 '12 at 10:38
It is true if $x≠kπ$. – Basil R Jun 3 '12 at 10:39
Not even. Take for example $\,x=\pi/4\,$ – DonAntonio Jun 3 '12 at 10:42
@DonAntonio: I took $k\in\mathbb Z$. But I am sorry I did not see he omit one sin($x)$ of denominator. – Basil R Jun 3 '12 at 10:50

You seem to have some serious problems with the algebra involved. Part of the problem is failure to use necessary parentheses: the result of applying the quotient rule is

$$\frac{2\sin x-(2x+1)\cos x}{\sin^2x}\;,$$

where the parentheses around $2x+1$ are absolutely necessary. If you choose to multiply out the numerator, you should get

$$\frac{2\sin x-2x\cos x-\cos x}{\sin^2x}\;.$$

Alternatively, you can split it into two fractions:

\begin{align*} \frac{2\sin x-(2x+1)\cos x}{\sin^2x}&=\frac{2\sin x}{\sin^2x}-\frac{(2x+1)\cos x}{\sin^2x}\\\\ &=2\csc x-(2x+1)\cot x\csc x\\\\ &=\csc x\Big(2-(2x+1)\cot x\Big)\;. \end{align*}

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You had (or should) $$\frac{2\sin x-2x\cos x-\cos x}{\sin^2x}$$ and this can't possibly equal what you wrote (where does that $\,3\,$ come from?)

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$$f(x)=\frac{2x+1}{\sin(x)}$$

As for the derivative of such a fractional function:

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

Simplifying:

$$f'(x)=\frac{2(\sin x)-\cos x(2x+1)}{\sin^2 x}=\frac{2\sin x-2x\cos x-\cos x}{\sin^2 x}=2\csc x-(2x+1) \cot x \csc x$$

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