Question

How do you find the derivative of `y=x^nlnx`?

Answer 1

`d/dx(x^nlnx)=x^(n-1) + nx^(n-1)lnx`

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^2sinxtanx` we have;

`{ ("Let "u = x^n, => , (du)/dx = nx^(n-1)'), ("And "v = lnx, =>, (dv)/dx = 1/x' ) :}`

Applying the product rule we get:

`\ \ \ \ \ \ \ \ \ \ \ d/dx(uv)=u(dv)/dx + (du)/dxv`
`:. d/dx(x^nlnx)=(x^n)(1/x) + (nx^(n-1))(lnx)`
`:. d/dx(x^nlnx)=x^(n-1) + nx^(n-1)lnx`