How do you differentiate #f(x)=x^3lnx # using the product rule?

1 Answer
Dec 12, 2016

Answer:

# d/dx(x^3lnx)=x^2 + 3x^2lnx #

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^3lnx # we have;

# { ("Let "u = x^3, => , (du)/dx = 3x^2'), ("And "v = lnx, =>, (dv)/dx = 1/x' ) :}#

Applying the product rule we get:

# \ \ \ \ \ \ \ \ \ \ \ d/dx(uv)=u(dv)/dx + (du)/dxv #
# :. d/dx(x^3lnx)=(x^3)(1/x) + (3x^2)(lnx) #
# :. d/dx(x^3lnx)=x^2 + 3x^2lnx #