How do you differentiate #y=x^2lnx#?

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Answer 1 · 2016-11-22T21:50:25

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

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^2lnx # we have;

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

# d/dx(uv)=u(dv)/dx + (du)/dxv #
# d/dx(x^2lnx)=(x^2)(1/x) + (2x)(lnx) #
# d/dx(x^2lnx)=x + 2xlnx #