how do you use logarithmic differentiation to find dy/dx?

Question

!) y=x^(2/x)
2)y=(x-2)^(x+1)

Answer 1

im not too much of a wordy person so i'll just solve the problem for u.

but here's the basic steps:
1. take the log of each side
2. "spread" them out as much as possible
3. differenciate each side, remember that for y. its gonna be (1/y) (dy/dx)
4. multiply each side by y.


1)y=x^(2/x)

ln y = ln x^(2/x) => ln y = 2/x ln x

(1/y) (dy/dx) = (2/x^2)(ln x ) + (2/x) (1/x)

dy/dx = (x^(2/x)) [(2 ln x / x^2) + (2/x^2)]


2) y = (x-2) ^ (x+1)

ln y = ln (x-2) ^ (x + 1)

ln y = (x+1) ln (x-2)

(1/y) (dy/dx) = (1) (ln (x -2) ) + (x+1) (1 / (x-2) )

dy/dx = [ (x-2) ^ (x+1)] [ ln (x-2) + (x+1) / (x-2) ]

Answer 2

y = x^(2/x)

Using logarithmic differentiation, take the ln of both sides.

ln(y) = ln(x^(2/x))

Use the log property that allows us to bring a power in front of the log.

ln(y) = (2/x) ln(x)

ln(y) = 2ln(x)/x

(1/y)(dy/dx) = 2[ (1/x)x - ln(x) ] / x^2
(1/y)(dy/dx) = 2[1 - ln(x)] / x^2

Multiply both sides by y,

dy/dx = 2y(1 - ln(x)) (1/x^2)

Plugging in y = x^(2/x), we get:

dy/dx = 2[x^(2/x)] (1 - ln(x)) (1/x^2)


2) y = (x - 2)^(x + 1)

ln(y) = (x + 1) ln(x - 2)

(1/y) (dy/dx) = ln(x - 2) + (x + 1)(1/(x - 2))

(1/y) (dy/dx) = ln(x - 2) + (x + 1)/(x - 2)

dy/dx = y [ln(x - 2) + (x + 1)/(x - 2)]

dy/dx = (x - 2)^(x + 1) [ln(x - 2) + (x + 1)/(x - 2)]

Answer 3

1) Take log of both sides:
log y = (2/x)logx
Differentiate:
(1/y)dy/dx=(2/x^2)-logx (2/x^2)

2) Similar for this.

Answer 4

1) y = x^(2/x)

Take the natural log of both sides.

ln(y) = ln[x^(2/x)] = (2/x)(ln x) = 2(ln x)/x

Take the derivative implicitly.

(1/y)(dy/dx) = 2[x(1/x) - (ln x)(1)]/x² = 2[1 - (ln x)]/x²
(dy/dx) = 2y[1 - (ln x)]/x²

Plug in the value of y.

(dy/dx) = 2x^(2/x)[1 - (ln x)]/x²
_____________________

2) y = (x - 2)^(x + 1)

Take the natural log of both sides.

(ln y) = ln[(x - 2)^(x + 1)] = (x + 1)ln(x - 2)

Take the derivative implicitly.

(1/y)(dy/dx) = ln(x - 2) + (x + 1)/(x - 2)
(dy/dx) = y[ln(x - 2) + (x + 1)/(x - 2)]

Plug in the value of y.

(dy/dx) = {(x - 2)^(x + 1)} [ln(x - 2) + (x + 1)/(x - 2)]