y=x^(9/x). What is the derivative?

Question

You might have to differntiate using natural ln.

Answer 1

(9/x) (x) ^ (1-9/x)

Answer 2

y = x ^ 9/x

y ' = ?

Solution: ln y = (9/x) (ln x)

y '/y = (9/x)' (ln x) + (ln x)' (9/x)

" = ((9)' (x) - (x)' (9))/x^2 (ln x) + (ln x)' (9/x)

" = -9 /x^2 (ln x) + (1/x) (9/x)

" = (-9 ln x)/x^2 + (9/x^2)

" = (-9 ln x + 9)/x^2

y '/y = (-9 ln x + 9)/x^2

y ' = x ^ 9/x (-9 ln x + 9)/x^2

y ' = 9x ^ 9/x (-ln x + 1)/x^2

Answer => y ' = 9x ^ 9/x ( 1 - ln x)/x^2

Answer 3

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Answer 4

just do this first...

ln y = 9/x * ln x

then differentiate both side. Implicit Differentiation.

Answer 5

y = x^(9/x)
= e^[(9lnx)/x] (since x = e^(lnx)
= e^u where u = 9lnx/x Thus du/dx = 9(1/x² - lnx)/x² by quotient rule
ie du/dx = 9(1 - x²lnx)/x^4

So dy/dx = dy/du * du/dx
= e^u * 9(1/x² - lnx)/x²
= 9(1 - x²lnx) / x^4 * x^(9/x)
= 9(1 - x²lnx)x^(9/x - 4)
= 9(1 - x²lnx)x^[(9 - 4x)/x]

PS Made some modifications on perusing ... sign and transcription errors .... sorry!!!

Answer 6

y = x^(9x^-1)

Take the ln of both sides

ln(y) = (9x^-1)ln(x)

Differentiate implicitly wrt x

1/y* dy/dx = -9x^-2*ln(x) + (9x^-1)*x^-1

dy/dx = y(-9)x^(-2)(1+ln(x))

Answer 7

take log on both sides
log y=9/x log x
xlogy=9logx
derivating wrt x
log y+x/y dy/dx =9/x
dy/dx=y/x(9/x-log y)

Answer 8

y'=(-9*(9/x)*x^(9/x-1))/x^2
or
lny=(9lnx)*x^-1
y'/y=9/x^2-(9lnx)/x^2
y'=y(9-9lnx)/x^2

yeah i think the natural log is easier