Differentiation foundations

The derivative of $4x^3$ is $12x^2$, that of $-5x$ is $-5$, and a constant has derivative zero. Therefore $\frac{dy}{dx}=12x^2-5$. The exponent drops by one after multiplying by its original value.

At $x=2$, $y=10$. Differentiating gives $y'=2x+3$, so the tangent’s gradient is $7$. Hence $y-10=7(x-2)$, or $y=7x-4$. Substituting $x=2$ into this line returns $y=10$ as required.

$y'=3x^2$ is positive for both $x<0$ and $x>0$. The curve is increasing on both sides, so the origin is neither a local maximum nor a local minimum. It is a stationary point of inflection. A zero derivative identifies a stationary point but does not, by itself, classify it.