What is the derivative of $y = x ln x$ $y= xlnx$ ?

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Answer 1 · 2017-05-16T17:33:21

$\frac{d y}{d x} = 1 + ln x$ $dy/dx = 1 + lnx$

We have:

$y = x ln x$ $y= xlnx$

We can apply the product rule to get:

$\frac{d y}{d x} = (x) (\frac{d}{d x} ln x) + (\frac{d}{d x} x) (ln x)$ $dy/dx = (x)(d/dx lnx) + (d/dxx)(lnx)$

Noting a standard calculus result:

$\frac{d}{d x} ln x = \frac{1}{x}$ $d/dx lnx = 1/x$

We get:

$\frac{d y}{d x} = x \cdot \frac{1}{x} + 1 \cdot ln x$ $dy/dx = x*1/x + 1*lnx$

$= 1 + ln x$ $" " = 1 + lnx$

Corollary
We have just shown that:

$\frac{d}{d x} (x ln x) = 1 + ln x$ $d/dx (xlnx) = 1 + ln x$

If we now integrate both sides, then we get:

$\ \ \ \ \ xlnx = int \ (1 + ln x) \ dx$

$:. xlnx = x -c + int \ ln x \ dx$

Hence:

$int \ ln x \ dx = xlnx -x +c$

What is the derivative of y= xlnxy=xlnx y= xlnx?