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)#