Question

Find dy/dx if y=6x²sinx cosx
​

Answer 1

`6\, (x^(2)\, \cos(2\, x) + x\, \sin(2\, x))`, or equivalently
`6\, (x^(2)\, \cos^(2)(x) - x^(2)\, \sin^(2)(x) + 2\, x \, \sin(x) \, \cos(x))`.

Explanation:

Make use of the double-angle identity of
`\sin(x)`:

`\sin(2\, x) = 2\, \sin(x) \, \cos(x)`.

`\begin{aligned} y &= 6\, x^(2)\, \sin(x)\, \cos(x) \\ &= 3\, x^(2)\, (2\, \sin(x)\, \cos(x)) \\ &= 3\, x^(2) \, \sin(2\, x)\end{aligned}`.

Apply the product rule
`(\ast)` and the chain rule
`(\ast \ast)`:

`\begin{aligned} & (d)/(dx)\left[3\, x^(2) \, \sin(2\, x)\right] \\ =\; & 3\, x^(2)\, (d)/(dx)[\sin(2\, x)] + (d)/(dx)\left[3\, x^(2)\right]\, \sin(2\, x) && (\ast) \\ =\; & 3\, x^(2)\, (2\, \cos(2\, x)) + 6\, x\, \sin(2\, x) && (\ast\ast) \\ =\; & 6\, x^(2)\, \cos(2\, x) + 6\, x\, \sin(2\, x)\end{aligned}`.

Optionally, rewrite
`\cos(2\, x)` and
`\sin(2\, x)` using double-angle identities:

`\begin{aligned} & (d)/(dx)\left[3\, x^(2) \, \sin(2\, x)\right] \\ =\; & \cdots \\ =\; & 6\, x^(2)\, \cos(2\, x) + 6\, x\, \sin(2\, x) \\ =\; & 6\, x^(2)\, (\cos^(2)(x) - \sin^(2)(x))+ 6\, x\, (2\, \sin(x)\, \cos(x)) \\ =\; & 6\, x^(2)\, \cos^(2)(x) - 6\, x^(2)\, \sin^(2)(x) + 12\, x\, \sin(x)\, \cos(x)\end{aligned}`.