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

##### 1 Answer

#### Answer:

# 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

For the first component Let

# 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

# 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 #