Question

2. Find the gradient of the curve y =4 sin 2x + 3cos 1/2x, at x = TT/3

Answer 1

Solution:

$y=4\sin 2x+3\cos (1/2\ x)$

On differentiating both sides w.r.t $x$,

$\frac{dy}{dx}=4\cos 2x.(2)+3(-\sin (\frac12 \ x)).(\frac12) \\=8\cos 2x-\frac32 (\sin \frac12x)$

Put $x=\frac{\pi}3$

$\frac{dy}{dx}=8\cos 2(\frac{\pi}3)-\frac32 (\sin \frac12.\frac{\pi}3) \\=8\cos (\frac{2\pi}3)-\frac32 (\sin \frac{\pi}6) \\=8(\frac{-1}{2})-\frac{3}{2}(\frac{1}{2}) \\=-4.75$

So, the gradient of $y$ at $x=\frac{\pi}3$ is $-4.75$.