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Find the derivative $\frac{d}{dx}\left(\frac{\sin\left(x\right)}{1+\cos\left(x\right)}+arctan\left(x\right)\ln\left(1+x^2\right)\right)$ using the sum rule

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Answer

$\frac{\cos\left(x\right)\left(1+\cos\left(x\right)\right)+\sin\left(x\right)^2}{\left(1+\cos\left(x\right)\right)^2}+\frac{1}{1+x^2}\ln\left(1+x^2\right)+x\frac{2}{1+x^2}arctan\left(x\right)$

Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{\sin\left(x\right)}{1+\cos\left(x\right)}+arctan\left(x\right)\cdot\ln\left(1+x^2\right)\right)$
1

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(\frac{\sin\left(x\right)}{1+\cos\left(x\right)}\right)+\frac{d}{dx}\left(arctan\left(x\right)\ln\left(1+x^2\right)\right)$
2

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=arctan\left(x\right)$ and $g=\ln\left(1+x^2\right)$

$\frac{d}{dx}\left(\frac{\sin\left(x\right)}{1+\cos\left(x\right)}\right)+\frac{d}{dx}\left(arctan\left(x\right)\right)\ln\left(1+x^2\right)+arctan\left(x\right)\frac{d}{dx}\left(\ln\left(1+x^2\right)\right)$

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Answer

$\frac{\cos\left(x\right)\left(1+\cos\left(x\right)\right)+\sin\left(x\right)^2}{\left(1+\cos\left(x\right)\right)^2}+\frac{1}{1+x^2}\ln\left(1+x^2\right)+x\frac{2}{1+x^2}arctan\left(x\right)$
$\frac{d}{dx}\left(\frac{\sin\left(x\right)}{1+\cos\left(x\right)}+arctan\left(x\right)\cdot\ln\left(1+x^2\right)\right)$

Main topic:

Sum rule

Used formulas:

7. See formulas

Time to solve it:

~ 0.89 seconds

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