# How do you find the gradient of the tangent to the curve y=x^3 at the given value of x=4?

You start by evaluating the derivative $y '$ of the function and then substitute $x = 4$ in it:
$y ' \left(x\right) = 3 {x}^{2}$
substitute $x = 4$;
$y ' \left(4\right) = 3 \cdot {4}^{2} = 48$ which is your gradient.