Derivative Of Ln 2x

What is the derivative of 2xlnx? What principle do you use for this? UC rules? please explain.

2xlnx results from two factors, 2x and lnx.

The derivative of 2x is 2.

The derivative of lnx is 1 / x.

So, according to the uct principle:

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

Rearrange the terms

= 2lnx + 2

The derivative of 2xlnx is 2 (1+ lnx). The steps are shown below.

We use the uct principle.

= 2 * (x) * derived from (ln x) + 2 * lnx * derived from (x)

= 2 * (x) * (1 / x) + 2 * lnx * (1)

= 2 * (1) + 2 * ln x

= 2 * (1 + lnx)

Is it a derivative or a necessary problem?

What I have described is related to the derivative, I think only you can move 2 and then run the formula. I'm sure you know w ... d (vu) = dvu + duv

If you merge ...

Integral after part, formula is F (u) d (v) = (u) (v) F (v)

Make u = lnx, du = dx / x, v = x 2, dv = 2x

Plug it in and you will get x 2lnxFx 2dx / x.

x 2lnx is enough in Fxdx.

For x 2lnx (x 2) / 2.

I thought it was just ...

Derivative Of Ln 2x

Derivative Of Ln 2x

(a * b) = ab + ba

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

Let f (x) = 2x.ln x

Then f (x) = 2 [x. (1 / x) + (lnx). (1)]

I am using the principle of uct differentiation. Who said yes

f (x) = g (x). Hour (x)

f (x) = g (x). j (x) + j (x). g (x)

uct rules

y = (2x). (LNX)

dy / dx = 2.lnx + (1 / x). (2x)

dy / dx = 2.lnx + 2

Derivative Of Ln 2x

Derivative Of Ln 2x

d / dx (2x lnx)

Let f (x) = 2x ln x

u = 2x> du / dx = 2

v = ln x> dv / dx = 1 / x

There f` (x) = v * du / dx + u * dv / dx (use Uct principle)

= (lnx) (2) + 2x (1 / x)

= 2 x x + 2

= 2 (1 + ln x) (years)

Derivative Of Ln 2x