Question

`\sf \large \: If \: y = Sin x \: * Cos \: 2x \: Find (dy)/(dx)`

Thanku!​

Answer 1

• cos (x) cos (2x) - 2sin(x) sin(2x)

Solution:

(See the solution in the photo)

Answer 2

`\sf -5\cos \left(x\right)+6\cos ^3\left(x\right)`

Step-by-step explanation:

`\sf y = sin(x) * cos(2x)`

`\rightarrow \sf (d)/(dx)\left(sin\left(x\right)\ * \:\:cos\left(2x\right)\right)`

`\sf \bold {Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'`

`\rightarrow \sf (d)/(dx)\left(\sin \left(x\right)\right)\cos \left(2x\right)+(d)/(dx)\left(\cos \left(2x\right)\right)\sin \left(x\right)`

`\sf \bold{ Apply \ differentiation \ rule \ \ \ : } \ \ \ sin(x) = cos(x) \ \ and \ \ cos(x) = -sin(x)`

`\rightarrow \sf \cos \left(x\right)\cos \left(2x\right)+\left(-\sin \left(2x\right)\ * \:2\right)\sin \left(x\right)`

`\rightarrow \sf \cos \left(x\right)\cos \left(2x\right)\left-2\sin \left(2x\right)\sin \left(x\right)`

`\sf \bold {use \ the \ formulae \ : \ cos(2x) = 2cos^2(x) - 1} \ {and} \ \ \sf \bold{sin(x) = 2 sin x cos x}`

`\rightarrow \sf cos(x) (2cos^2 (x) -1) -2(2sin(x)cos(x)sin(x))`

`\rightarrow \sf 2cos^3 (x) - cos(x) - 4sin^2(x) cos(x)`

`\rightarrow \sf -5\cos \left(x\right)+6\cos ^3\left(x\right)`