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What is the derivative of ##y=arctan(4x)##?
Answer ##4/(16x^2 + 1)##
Explanation First recall that ##d/dx[arctan x] = 1/(x^2 + 1)##.
Via the :
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 voilĂ - there's our identity.