Use logarithmic differentiation to find dy/dx.

Question

y ¯= x² √x-7 , x > 7

Answer 1

ln y = ln [x^2 * sqrt(x-7)]

= 2 ln x + (1/2) ln (x-7)

y' / y = 2 / x + (1/2) ( 1/(x-7))

y' = x^2 * sqrt(x-7) [ 2/ x + (1/2) (1/(x-7))

Answer 2

So if we have

y(x) = x^2 * Sqrt[x - 7]

then, taking the natural log, we find:

ln(y) = ln(x^2 * Sqrt[x-7])

= ln(x^2) + ln((x-7)^1/2)

= 2 * ln(x) + (1/2) * ln(x-7)

If we take the derivative of both sides (noting that d/dx ln(y(x)) = y'(x) / y(x)) we find

y'(x) / y(x) = 2/x + 1 / (2 * (x-7))

= (4 * (x - 7) + x) / (2 * x * (x - 7))

= (5 * x - 28) / (2 * x * (x - 7))

Multiplying both sides by y(x), we find:

y'(x) = x^2 * Sqrt[x-7] * (5 * x - 28) / (2 * x * (x - 7))

= x * (5 * x - 28) / (2 * Sqrt[x - 7])