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- Anonymous2 years ago
y=In(x^2+5)

d y / d x = [ 2 x / (x^2+5) ]

- VamanLv 72 years ago
y=In(x2+5)

This gives x2+5 = exp y

Take the derivative. 2x = exp y dy/dx

Therefore dy/dx = 2x/exp y= 2x/(x^2 +5)

- ComoLv 72 years ago
y = ln ( x² + 5 )

Let u = x² + 5

du/dx = 2x

y = ln u

dy/du = 1/u

dy/dx = (1/u) (2x) = 2x / ( x² + 5 )

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- L. E. GantLv 72 years ago
assuming you mean y = ln(x^2 + 5)...

Let u = x^2 + 5

so y = ln(u)

dy/du = 1/u

du/dx = 2x

by the chain rule dy/dx = dy/du * du/dy = 2x/(x^2 + 5)

- Anonymous2 years ago
Use chain rule

u = x^2 + 5

du/dx = 2x

dy/du = ln(u)

dy/du = 1/u

dy/dx = dy/du * du/dx

dy/dx = 1/u * 2x

dy/dx = 2x / (x^2+5)

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