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Your question is pretty ambiguous. First of all, the natural logarithm is a function and requires a variable to be input into it. Secondly, the process of differentiation is a type of transformation or operation performed on a function, and it is done with respect to a particular variable. Its important when you ask questions in math that you ask a full, explicit question.
"How do you differentiate, with respect to x, the function ln(x)" would be more proper.
Or simply do it in math:
d/dx ln(x)
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The answer to the question is simple. It is 1/x.
d/dx ln(x) = 1/x
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Differentiate y equals a power x?
y=ax y'=ln(a)*ax
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)
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
How do you Differentiate and Integrate y equals 3 to the power of x?
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
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))
