How do you differentiate #arctan(x^2)#?

2 Answers
mason m
Jun 6, 2016

#(2x)/(1+x^4)#

Explanation:

The derivative of the arctangent function is:

#d/dxarctan(x)=1/(1+x^2)#

So, when applying the chain rule, this becomes

#d/dxarctan(f(x))=1/(1+(f(x))^2)*f'(x)#

So, for #arctan(x^2)#, where #f(x)=x^2#, we have

#d/dxarctan(x^2)=1/(1+(x^2)^2)*d/dx(x^2)#

#=(2x)/(1+x^4)#

mason m
Jun 6, 2016

#(2x)/sqrt(1+x^4)#

Explanation:

Let #y=arctan(x^2)#.

Then #tan(y)=x^2#. From here, differentiate both sides of the equation. Recall that the chain rule comes into effect on the left-hand side.

#sec^2(y)*dy/dx=2x#

Dividing both sides by #sec^2(y)#, which is equivalent to multiplying both sides by #cos^2(y)#, gives

#dy/dx=cos^2(y)*2x#

#dy/dx=cos^2(arctan(x^2))*2x#

Note that #cos^2(arctan(x^2))# can be simplified.

If #arctan(x^2)# is an angle in a right triangle, then #x^2# is the side opposite the angle and #1# is the side adjacent to the angle. Then, by the Pythagorean Theorem, #sqrt(1+x^4)# is the triangle's hypotenuse.

Since cosine is the adjacent side, #1#, divided by the hypotenuse, #sqrt(1+x^4)#, we see that #cos(arctan(x^2))=1/sqrt(1+x^4)#.

Therefore

#dy/dx=(1/sqrt(1+x^4))^2*2x=1/(1+x^4)*2x=(2x)/(1+x^4)#

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