what is the derivative of 2xlnx. What rule do you use for this? the product rule? Plz explain.?

Answer 1

2xlnx is a product of two factors, 2x and lnx
the derivative of 2x is 2
the derivative of lnx is 1/x

so using the product rule:

d(u*v) = u * dv + v * du
let u = 2 x and v = ln x

d(2x*lnx)/dx
= lnx * d(2x)/dx + 2x * d(lnx)/dx
=lnx * 2 + 2x* 1/x
rearranging terms
= 2lnx + 2

Answer 2

the derivative of 2xlnx is 2(1+ lnx). the steps are given below.
we use product rule.
= 2 * (x) * derivative of (ln x) + 2 * lnx * derivative of (x)
= 2 * (x) * (1/x) + 2 * ln x * (1)
= 2 * (1) + 2*ln x
= 2 * (1 + ln x)

Answer 3

is this a derivative or integral problem?
what i explain is intergral.. with derivative, i think its just product you can move the 2 out and then do the formula, im sure u know how... d(vu) = dvu+duv

if you intergral...
integral by parts, the formula is F(u)d(v) = (u)(v)- F(v)du

make u=lnx, du=dx/x, v=x^2, dv=2x

plug in and you get x^2lnx-Fx^2dx/x
simply to x^2lnx- Fxdx
that become x^2lnx- (x^2)/2
i think this is right...

Answer 4

(a*b)' = a'b + b'a
ln'(x) = 1/x
x' = 1

(2xlnx)' = 2(x*ln x)' = 2*(x' lnx + x ln' x) = 2 (ln(x)+ x*(1/x)) = 2lnx + 2

Answer 5

say f(x) = 2x.ln x
then f'(x) = 2[x.(1/x) + (ln x).(1)]

I used the product rule of differentiation. which states that if
f(x) = g(x) . h(x)
f'(x) = g(x) . h'(x) + h(x) . g'(x)

Answer 6

Product rule
y = (2x).(ln x)
dy/dx = 2.ln x + (1/x).(2x)
dy/dx = 2.lnx + 2

Answer 7

d/dx(2x ln x)

let f(x) = 2x ln x
u= 2x -->du/dx=2
v= ln x-->dv/dx=1/x
therefore f`(x)= v*du/dx+ u*dv/dx (use product rule)
= (ln x)(2) + 2x(1/x)
= 2 ln x + 2
= 2( 1 + ln x) (ans)