How do you find the derivative of the function `y= tan^-1 (3x)`?

Remembering that the derivative of:

`y=tan^-1x`

is

`y'=1/(1+x^2)`

than:

`y'=1/(1+(3x)^2)*3=(3)/(1+9x^2)`.

By the Chain Rule (and knowledge of the derivative of inverse tangent function), the answer is: `dy/dx=\frac{1}{1+(3x)^{2})\cdot 3=\frac{3}{1+9x^{2}}`.

If you didn't happen to remember the derivative of the inverse tangent function, you could derive the answer by differentiation of both sides of the equation `tan(tan^{-1}(3x))=3x` with respect to `x` to get, by the Chain Rule, `sec^{2}(tan^{-1)(3x))\cdot \frac{d}{dx}(tan^{-1}(3x))=3` so that `\frac{d}{dx}(tan^{-1}(3x))=3cos^{2}(tan^{-1}(3x))`.

To simplify this, at this point, you could draw a right triangle and label one of the non-right angles `\tan^{-1}(3x)`. Since the tangent of the angle is the length of the opposite side divided by the length of the adjacent side, you could label the opposite side with "`3x`" and the adjacent side with "1". The Pythagorean Theorem would imply that the length of the hypotenuse is `\sqrt{1+9x^{2}}`. Since the cosine of the angle is the length of the adjacent side divided by the length of the hypotenuse, you'd get `\cos(\tan^{-1}(3x))=\frac{1}{\sqrt{1+9x^{2}}`.

Hence, `\frac{d}{dx}(tan^{-1}(3x))=3cos^{2}(tan^{-1}(3x))=\frac{3}{1+9x^{2}}`