How do you differentiate $y = x^{2} ln x$ $y=x^2lnx$ ?

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Answer 1 · 2016-11-22T21:50:25

$\frac{d}{d x} (x^{2} ln x) = x + 2 x ln x$ $d/dx(x^2lnx)=x + 2xlnx$

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

$\frac{d}{d x} (u v) = u \frac{d v}{d x} + \frac{d u}{d x} v$ $d/dx(uv)=u(dv)/dx+(du)/dxv$ , 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;

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

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

How do you differentiate y=x^2lnxy=x2lnxy=x^2lnx?