Answer to Question #188141 in Calculus for VPM

Question #188141

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


1
Expert's answer
2021-05-07T10:10:30-0400

Solution:

y=4sin2x+3cos(1/2 x)y=4\sin 2x+3\cos (1/2\ x)

On differentiating both sides w.r.t xx,

dydx=4cos2x.(2)+3(sin(12 x)).(12)=8cos2x32(sin12x)\frac{dy}{dx}=4\cos 2x.(2)+3(-\sin (\frac12 \ x)).(\frac12) \\=8\cos 2x-\frac32 (\sin \frac12x)

Put x=π3x=\frac{\pi}3

dydx=8cos2(π3)32(sin12.π3)=8cos(2π3)32(sinπ6)=8(12)32(12)=4.75\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 yy at x=π3x=\frac{\pi}3 is 4.75-4.75.


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