If for x(0,14), the derivative of ^-1(6x x1-9x^3) is x g(x), then g(x) equals:

Question

If for $x\in\left(0,\dfrac14ight)$, the derivative of $	an^{-1}\left(\dfrac{6x\sqrt x}{1-9x^3}ight)$ is $\sqrt x\cdot g(x)$, then $g(x)$ equals:

Accepted Answer

$F(x)=\tan^{-1}\left(\dfrac{2\cdot3x^{3/2}}{1-(3x^{3/2})^2}\right)=2\tan^{-1}(3x^{3/2})$.

$$F'(x)=\frac{2}{1+9x^3}\cdot3\cdot\frac32x^{1/2}=\frac{9\sqrt x}{1+9x^3}$$

So $g(x)=\dfrac{9}{1+9x^3}$.

Solutions · 1