How do you differentiate (In x)^2?

Answer 1

Let y = (In x)^2. This is a function of a function.

Use the chain rule to find dy/dx

Let In x = u, then y = u^2.

dy/dx = 2u = 2 In x = Inx^2

and du/dx = 1/x.

Hence dy/dx = dy/du times du/dx = Inx^2 times 1/x

= (Inx^2)/(x) *brackets added for clarity.

Answer 2

Differentiate Ln X 2

Answer 3

I think you'd have to use the chain rule. Differentiate ln x, which is 1/x, and then suppose your other function is x^2. Thus, dy/dx = (dy/du)(du/dx). Thus, do 2(ln x)(1/x). I think....

Answer 4

For this problem you use the chain rule.

the answer is 2*ln x * (1/ x) = 2(ln x) / x.

good luck

Answer 5

y=[lnx]^2
y'=2lnx*1/x
=2 lnx/x

Answer 6

Answer = (2/x) ln(x)

Answer 7

dy/dx=xd/dx2^(x)+ 2^(x) d/dx(x) = x.2^(x).log2 +2^(x) therefore dy/dx =2^(x)[xlog2+1]

Answer 8

here we use the chain rule

d(uv)/dx=(u)dv/dx+(v)ud/dx.(1)

let y=(lnx)^2, =lnx*lnx, u=lnx,v=lnx

substitute into (1)

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

=(2lnx)/x

i hope that this helps

Answer 9

(ln(x))^2 = (ln(x)) * (ln(x))

(ln(x)')(ln(x)) + (ln(x))(ln(x))'

(1/x)(ln(x)) + (ln(x))(1/x)

(ln(x)/x) + (ln(x)/x)

(2ln(x))/x or (ln(x^2))/x