Question

How do you differentiate `f(x)=x^2sinxtanx`?

Answer 1

`f'(x)=2xsinxtanx + x^2cosxtanx + x^2sinxsec^2x`

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

So with `f(x)=x^2sinxtanx` we have;

`{ ("Let "u=x^2, => , (du)/dx=2x), ("And "v=sinx, =>, (dv)/dx=cosx ), ("And "w=tanx, =>, (dw)/dx=sec^x ) :}`

`f'(x) = (du)/dxvw + u(dv)/dxw+ + uv(dw)/dx`
`f'(x)=(2x)(sinx)(tanx) + (x^2)(cosx)(tanx) + (x^2)(sinx)(sec^2x)`
`f'(x)=2xsinxtanx + x^2cosxtanx + x^2sinxsec^2x`