What is the derivative of # (arctan x)^3#?

Question

Luke Alvarez · Accepted Answer

#dy/dx=(3arctan^2(x))/(x^2+1)# We are given #y=arctan^3(x)# and need to find #dy/dx#. By the power rule, if #color(red)(y=x^a)#, then, #color(red)(dy/dx=ax^(a-1))#. Unfortunately, we can't directly apply the power rule to find #dy/dx# here as the base is #arctan(x)#, not #x#. Instead, the power rule can only find the derivative with respect to #arctan(x)#, or #dy/(d(arctan(x)))=3arctan^2(x)#. However, by the chain rule, we can multiply both sides by #color(red)((d(arctan(x)))/dx)# to get #dy/dx=3arctan^2(x)(d(arctan(x)))/dx#. (Note: the left hand side cancels out: #dy/cancel(d(arctan(x)))*cancel(d(arctan(x)))/dx=dy/dx#.) Now, we just need to find #(d(arctan(x)))/dx#. Consider the function #u=arctan(x)#. This necessarily means that #tan(u)=x#. Now, if we differentiate both sides, we will get #1/cos^2(u)=dx/(du)#, or #(du)/dx=cos^2(u)#. We said previously that #u=arctan(x)#. Substitute this in to find that #(du)/dx=cos^2(arctan(x))=1/(tan^2(arctan(x))+1)=1/(x^2+1)#. Note: #cos^2(arctan(x))# is simplified using the identity #tan^2(theta)+1=1/cos^2(theta)# (found by dividing both sides of #sin^2(theta)+cos^2(theta)=1# by #cos^2(theta)# and using the fact that #tan(theta)=sin(theta)/cos(theta)#). This identity can be arranged to #cos^2(theta)=1/(tan^2(theta)+1)#. Now, we have #(d(arctan(x)))/dx=1/(x^2+1)#. Since #dy/dx=3arctan^2(x)(d(arctan(x)))/dx#, our final answer is #dy/dx=(3arctan^2(x))/(x^2+1)#.

Luke Alvarez · Answer

undefined The derivative of $ (\arctan(x))^3 $ is $ \frac{3(\arctan(x))^2}{1+x^2} $.