Can you please explain how to differentiate y=ln(5xxx-2xx+1)

You would simply use the chain rule.

Moreover dy/dx (ln u) = [1/u][du/dx]


dy...........1
—..=..—————— times 15x² - 4x
dx.......5x³ -2x² + 1

we can write that as one rational expression



dy........15x² - 4x
—. =.——————
dx......5x³ -2x² + 1

i think you use chain rule but I dont remeber if its the derivative of the outside x the inside or vise vera. However, I am more inclined to say that you want to take the derivative of the inside x the outside. I cant remember the rules for natural logs though so keep them in mind.
I think it is the
y=ln(5x^3 - 2x^2 +1)
y=ln(15x^2 - 4x)
Sorry i dont remeber the rules for ln

y = ln ( 5 x ³ - 2 x ² + 1 )

y = ln u

u = 5 x ³ - 2 x ² + 1
du/dx = 15 x ² - 4x

dy/du = 1/u

dy/dx = (1/u) ( 15 x ² - 4 x )

dy/dx = (15 x ² - 4x) / ( 5 x ³ - 2 x ² + 1 )

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