How do you differentiate $f(x)=(1+arctanx)/(2-3arctanx)$ ?

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Answer 1 · 2017-03-24T05:21:19

$(df(x))/(dx)=5/((1+x^2)(2-3arctanx)^2)$

Here we can use quotient rule.

Also note that derivative of $arctanx$ , which is also known as $tan^(-1)x$ is $1/(1+x^2)$ i.e.

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

Now according to quotient rule if $f(x)=(g(x))/(h(x))$

then $(df(x))/(dx)=(h(x)*(dg(x))/(dx)-g(x)*(dh(x))/(dx))/((h(x))^2)$

Here $g(x)=1+arctanx$ hence $(dg(x))/(dx)=1/(1+x^2)$

and $h(x)=2-3arctanx$ hence $(dh(x))/(dx)=-3/(1+x^2)$

Hence

$(df(x))/(dx)=((2-3arctanx)*1/(1+x^2)-(1+arctanx)(-3/(1+x^2)))/((2-3arctanx)^2)$

= $((2/(1+x^2)-(3arctanx)/(1+x^2)+3/(1+x^2)+(3arctanx)/(1+x^2)))/((2-3arctanx)^2)$

= $(5/(1+x^2))/((2-3arctanx)^2)$

= $5/((1+x^2)(2-3arctanx)^2)$

How do you differentiate f(x)=(1+arctanx)/(2-3arctanx)f(x)=(1+arctanx)/(2-3arctanx)?