What is the derivative of $x ln(arctan x)$ ?

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Answer 1 · 2018-03-17T23:09:40

The derivative is $=ln(arctan(x))+x/((1+x^2))*1/(arctan(x))$

Let $u=arctanx$

Then,

$tanu=x$

$sec^2u=1+x^2$

Differentiating wrt $x$

$(du/dx)*sec^2u=1$

$(du)/dx=1/sec^2u=1/(1+x^2)$

Therefore,

$y=xln(arctanx)$

$y=xln(u)$

Differentiating wtt $x$ by pplying the product rule

$dy/dx=1*lnu+x*(du)/dx*1/u$

$=ln(arctan(x))+x/((1+x^2))*1/(arctan(x))$

Answer 2 · 2018-03-17T23:12:33

$d/dxxlnarctanx=lnarctanx+x/((1+x^2)arctanx)$

To find $d/dxxlnarctanx$ , we use the product rule

$d/dxuv=v(du)/dx+u(dv)/dx$

$d/dxlnarctanx=1/((1+x^2)arctanx)$

So

$d/dxxlnarctanx=lnarctanx+x/((1+x^2)arctanx)$

What is the derivative of x ln(arctan x) x ln(arctan x)?