Find the equation of the tangent to the curve with equation $y = 2x^{3} - 3x$ at the point where $x = 1$. - Scottish Highers Maths - Question 2 - 2023

Next, we need to differentiate the equation $y = 2x^{3} - 3x$ to find the gradient of the tangent line: $\frac{dy}{dx} = 6x^{2} - 3.$

Now, we evaluate the derivative at $x = 1$ to find the gradient: $\frac{dy}{dx}\bigg|_{x=1} = 6(1)^{2} - 3 = 6 - 3 = 3.$
Therefore, the gradient is $3$.

We can use the point-slope form of a linear equation, which is given by: $y - y_1 = m(x - x_1),$
where $(x_1, y_1) = (1, -1)$ and $m = 3$: $y - (-1) = 3(x - 1).$
This simplifies to: $y + 1 = 3x - 3,$ which can be rearranged to: $y = 3x - 4.$
Thus, the equation of the tangent line is $y = 3x - 4$.