Derivative of arctan. What is the derivative of the arctangent function of x? The derivative of the arctangent function of x is equal to 1 divided by (1+x 2) Integral of arctan. What is the integral of the arctangent function of x? The indefinite integral of the arctangent function of x is.
1 Answer
Jul 31, 2014
Answer
#4/(16x^2 + 1)#
#4/(16x^2 + 1)#
Explanation
First recall that #d/dx[arctan x] = 1/(x^2 + 1)#.
First recall that #d/dx[arctan x] = 1/(x^2 + 1)#.
Via the chain rule:
1.) #d/dx[arctan 4x] = 4/((4x)^2 + 1)#
2.) #d/dx[arctan 4x] = 4/(16x^2 + 1)#
If it isn't clear why #d/dx[arctan x] = 1/(x^2 + 1)#, continue reading, as I'll walk through proving the identity.
We will begin simply with
1.) #y = arctan x#.
From this it is implied that
2.) #tan y = x#.
Using implicit differentiation, taking care to use the chain rule on #tan y#, we arrive at:
3.) #sec^2 y dy/dx = 1#
Solving for #dy/dx# gives us:
4.) #dy/dx = 1/(sec^2 y)#
Which further simplifies to:
5.) #dy/dx = cos^2 y#
Next, a substitution using our initial equation will give us:
6.) #dy/dx = cos^2(arctan x)#
This might not look too helpful, but there is a trigonometric identity that can help us.
Recall #tan^2alpha + 1 = sec^2alpha#. This looks very similar to what we have in step 6. In fact, if we replace #alpha# with #arctan x#, and rewrite the #sec# in terms of #cos# then we obtain something pretty useful:
#tan^2(arctan x) + 1 = 1/(cos^2(arctan x))#
This simplifies to:
#x^2 + 1 = 1/(cos^2(arctan x))#
Now, simply multiply a few things around, and we get:
#1/(x^2 + 1) = cos^2(arctan x)#
Beautiful. Now we can simply substitute into the equation we have in step 6:
7.) #dy/dx = 1/(x^2 + 1)#
And voilà - there's our identity.
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