Answer to Question #158956 in Calculus for Daniel Howells

Question

y=3x³+2x²+4x+27

Answer 1

Here we have, $y=3x^3+2x^2+4x+27$

Now, we differentiate both sides wrt x:-

$y'=\frac{dy}{dx}=9x^2+4x+4$

Hence we have:-

$$\color{blue} y'=9x^2+4x+4$$