Question

How do you differentiate `y=x^2sinx+xcosx`?

Answer 1

`dy/dx = x^2cosx + xsinx + cosx`

Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

`d/dx(uv)=u(dv)/dx+(du)/dxv`, or, `(uv)' = (du)v + u(dv)`

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

`d/dx(uvw)=uv(dw)/dx+u(dv)/dxw + (du)/dxvw`

So with `f(x) = x^2sinx + xcosx` we will need to apply the product rule twice;

For the first component Let `y = x^2sinx`
`{ ("Let "u = x^2, => , (du)/dx = 2x), ("And "v = sinx, =>, (dv)/dx = cosx ) :}`

`d/dx(uv)=u(dv)/dx + (du)/dxv`
`:. d/dx(x^2sinx)=(x^2)(cosx) + (2x)(sinx)`
`:. d/dx(x^2sinx)=x^2cosx + 2xsinx` ..... [1]

For the second component Let `y = xcosx`
`{ ("Let "u = x, => , (du)/dx = 1), ("And "v = cosx, =>, (dv)/dx = -sinx ) :}`

`d/dx(uv)=u(dv)/dx + (du)/dxv`
`:. d/dx(xcosx)=(x)(-sinx) + (1)(cosx)`
`:. d/dx(xcosx)=cosx - xsinx` ..... [2]

Combining the results [1] ad [2] we get;

`dy/dx = x^2cosx + 2xsinx + cosx - xsinx`
`:. dy/dx = x^2cosx + xsinx + cosx`