Differentiation of sin and cos

Differentiating $y = \sin (x)$

In the diagram below, the curve

y = \sin (x)

is sketched. At the same time, the traced point represents the value of the gradient of $\sin (x) .$

Do you recognise the curve formed by the value of the gradient? It is $\cos (x) .$ Interesting!

This suggests that

y = \sin (x) \Rightarrow \frac{d y}{d x} = \cos (x)

If we do similar for $y = \cos (x),$ we find the following:

The gradient of the curve appears to give $- \sin (x),$ which suggests

y = \cos (x) \Rightarrow \frac{d y}{d x} = - \sin (x)

Find the gradient of the curve

y = 2 \sin (x) - 3 \cos (x)

at the point where

x = \frac{π}{6}