derivative of xlnx ?

Answer 1

Use the product rule

d(x)lnx + x*d(ln(x)) = ln(x) + x/x = ln(x) + 1

Answer 2

In order to do this, you have to use the product rule.
The product rule verbally goes as follows: "The derivative of the first times the second plus the derivative of the second times the first."

Note that the derivative of x is 1, and the derivative of lnx is 1/x. Therefore, if

f(x) = x lnx, then
f'(x) = (1) lnx + (1/x)(x)
f'(x) = lnx + 1

Answer 3

Write f(x) = xlnx then the derivative of cos(f(x)) is what you are looking for. d/dx cos(f(x)) = f'(x) * -sin(f(x), the standard way to differentiate cos. Use the product rule on f to find its derivative: f'(x) = d/dx(x) * lnx + d/dx(lnx) * x = lnx + 1 So the derivative is just -(lnx + 1)*sin(xlnx), not so scary after all.

Answer 4

d (xlnx) = lnx + 1

Answer 5

I just finished calc I and this is how it goes:
let "= prime or derivaite of y
y=xlnx

you have to use product rule.
y"=1*lnx+x(1/x)


y"=lnx+1

Answer 6

use product rule
x'lnx + x(lnx)'
= lnx + 1

Answer 7

d/dx f(x)*g(x) = f(x)*d/dx g(x) + g(x)*d/dx f(x). Substitute
x*d/dx lnx + lnx d/dx x evaluate
x* 1/x + lnx * 1 simplify
1 + lnx (or, equivelently, lnx + 1)


Doug

Answer 8

f(x) = xln(x)

f'(x) = x'ln(x) + xln(x)'
f'(x) = 1*ln(x) + x(1/x)
f'(x) = ln(x) + 1