Find derivative of y using logarithmic differentiation?

y = [ 1 + (6/x) ]^(8x)

take logs

ln y = 8x ln [1 + (6/x) ]

differentiate

y ' /y = 8x /(1 + 6/x)* (- 6/x^2) + 8 ln[1 + (6/x)]

y ' = y [ 8x^2 /(x + 6)*(-6/x^2) + 8ln [ 1 + (6/x)]

y ' = [1 + (6/x)]^8x [ - 48 /(x + 6) + 8ln [ 1 + (6/x) ]

y ' = 8[1 + (6/x)^8x ][ln [ 1 + (6/x) - 6/(x + 6) ]

ln(y) = 8x ln (1+ 6/x)

differentiate both sides with respect to x

Apply the product rule on the right side (with chain rule)

(1/y) dy/dx = 8 ln (1+6/x) + 8x / (1+6/x) d/dx(6/x)

(1/y) dy/dx = 8 ln (1+6/x) + [8x /(1+6/x)] [ -6/x^2]

(1/y) dy/dx = 8 ln (1+6/x) - 48 /x(1+6/x)

(1/y) dy/dx = 8 ln (1+6/x) - 48 /(x+6)

dy/dx = y { 8 ln (1+6/x) - 48 /(x+6) }

dy/dx = 1+(6/x))^(8x) { 8 ln (1+6/x) - 48 /(x+6) }

First set up the concern. y = x^(a million/x) Then take the organic log of the two area. ln y = ln (x^(a million/x)), and simplifying, taking the exponent out of the log, we get ln y = ln(x)/x Then, take the by-manufactured from the two area, with admire to x. The by-manufactured from the left area, d(ln y)/dx = (a million/y)*(y'). the y' (comparable as dy/dx) is there via chain rule. For the marvelous area, we would desire to apply the quotient rule. for 2 function g and f, (f/g)' = (gf' - fg')/g^2 So our f subsequently is f = ln(x) and g = x subsequently (ln(x)/x)' = (x(a million/x) - (ln(x)*(-a million/x^2)))/(x^2) So we've (a million/y)*(y') = (x(a million/x) - (ln(x)*(-a million/x^2)))/(x^2) Multiply by y on the two area and get y' = (y) * (x(a million/x) - (ln(x)*(-a million/x^2)))/(x^2) all of us understand that y = x^(a million/x), so substitute that throughout for (y) And we get y' = x^(a million/x) * (x(a million/x) - (ln(x)*(-a million/x^2)))/(x^2)