Question

d/dx of y=arctan\sqrt(6x)

Answer 1

Refer to the attachment, a solution using the Mathematica computer program.

Answer 2

Note
$$
\cfrac{d}{dx}\tan x=\cfrac{1}{1+x^2}
$$
Because
$$
\text{Letting x=tan y} \text{ then because }\cfrac{d}{dy}\tan y=\sec^2 x \text{ and } 1+\tan^2x=\sec^2x,\\\\
\int \cfrac{1}{1+x^2}dx=\int \cfrac{\sec^2y}{1+\tan^2y}dy=\int dy=y=\tan^{-1} x\\
$$
Using this fact and the product rule,
$$
\cfrac{d}{dx}\tan^{-1}y=\cfrac{1}{1+y^2}\cfrac{dy}{dx}
$$
So for your problem, \(y=\sqrt {6x}\):
$$
\cfrac{d}{dx} \tan^{-1}\sqrt{6x}=\frac{\cfrac{d}{dx}\sqrt{6x}}{1+(\sqrt{6x})^2}
$$
Which you should be able to evaluate.

Answer 3

i would go down the \(\tan y = \sqrt{6x}\) route and do it implicitly...

or maybe even quicker, write it as \(f(x,y) = \tan y - \sqrt{6x} = 0\) so \(y' = -\dfrac{f_x}{f_y} = -\dfrac{-{\sqrt{6}\over 2}x^{-1/2}}{\sec^2 y}\) and from \(\tan^2 y + 1 = \sec^2 y\) we have \(y' = \dfrac{\sqrt{{3}\over 2}x^{-1/2}}{1+6x} = \sqrt{{3}\over 2}\dfrac{1}{\sqrt{x} (1+6x)}\)