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Let y = ln(x2), where ln is log to base e.
Using our log laws, y = 2 ln (x)
Therefore dy/dx = 2 . 1/x = 2/x
Another way to do this is to use the chain rule (for differentiation)
Let y = ln(x2), where ln is log to base e.
Let u = x2, therefore du/dx = 2x
y = ln u, therefore dy/du = 1/u
Putting it altogether gives:
dy/dx = dy/du . du/dx = 1/u . 2x = 2x/x2 = 2/x
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Is logarithmic differentiation specially useful when dealing with functions with exponent that also depend on x?
Yes. For example, to differentiate y = (x^2 + 1)^x, we take the natural log of both sides.ln(y) = ln((x^2 + 1)^x) Bring down the exponent. ln(y) = x ln(x^2 + 1) Differentiate both sides. dy/y = ((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Substitute in y = (x^2 + 1)^x. dy/((x^2 + 1)^x) =((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Solve for dy/dx. dy/dx = ((x^2 + 1)^x)((2x^2)/(x^2 + 1) + ln(x^2 + 1))
What is the antiderivative of e to the power of one divided by x?
1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace
What is the anti-derivative of 5 to the x power?
ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)
Solve or x 2 ln 9 plus 2 ln 5 equals 2 ln x minus 3?
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
Differentiate y equals a power x?
y=ax y'=ln(a)*ax
