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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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    @DonAntonio: I took $k\in\mathbb Z$. But I am sorry I did not see he omit one sin($x)$ of denominator.2012-06-03

3 Answers 3

3

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

As for the derivative of such a fractional function:

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

Simplifying:

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

4

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?)

3

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:

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