# Determine the first derivative (dy/dx) of y= tan (3x^2)

Determine the first derivative

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Determine the first derivative
Let $u=3{x}^{2}$, then
$y=\mathrm{tan}\left(u\right)$
And, $\frac{dy}{dx}=\left(\frac{dy}{du}\right)\cdot \left(\frac{du}{dx}\right)$
Taking $y=\mathrm{tan}\left(u\right)$, then $\frac{dy}{du}={\mathrm{sec}}^{2}\left(u\right)$
Taking $u=3{x}^{2}$, then $\frac{du}{dx}=6x$
So, $\frac{dy}{dx}=\left(\frac{dy}{du}\right)\cdot \left(\frac{du}{dx}\right)$
$\frac{dy}{dx}=\left({\mathrm{sec}}^{2}\left(u\right)\right)\cdot \left(6x\right)$
$\frac{dy}{dx}=6x\cdot {\mathrm{sec}}^{2}\left(3{x}^{2}\right)$