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How do you differentiate ln?

Anonymous

∙ 15y ago

Updated: 10/24/2022

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)

============

The answer to the question is simple. It is 1/x.
d/dx ln(x) = 1/x

Wiki User

∙ 15y ago

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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))

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Related Questions

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)

How do you differentiate ln1.01?

The derivative of the ln function is the function 1/x. So the derivative of ln1.01 should be 1/1.01 = 0.990099... ------------------------- Well I may be looking at this slightly different, but the question as stated "differentiate ln(1.01)" would be 0 seeing as ln(1.01) is itself a constant (irrational) number. The derivative of any constant is zero. If the intended question was ln(x)d/dx where x=1.01 then I agree with the above answer.

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).

How do you differentiate exponential function?

The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)

What is the derivative of lnx raised to lnx?

start by setting y=lnx^lnx take ln of both sides lny=lnx(ln(lnx)) differentiate dy/dx(1/y)=(1+ln(lnx))/x dy/dx=y(1+ln(lnx))/x we know that y=lnx^lnx so we can just substatute back in dy/dx=(lnx^lnx)*(1+ln(lnx))/x

How do you differentiate 2 to the power x with respect to x?

The answer is ln(2)2x where ln(2) is the natural log of 2. The answer is NOT f(x) = x times 2 to the power(x-1). That rule applies only when the exponent is a constant.

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))

How would you solve ln 4 plus 3 ln x equals 5 ln 2?

Ln 4 + 3Ln x = 5Ln 2 Ln 4 + Ln x3= Ln 25 = Ln 32 Ln x3= Ln 32 - Ln 4 = Ln (32/4) = Ln 8= Ln 2

How do you differentiate ln 2 to the exponent x?

ln2^x = xln2. let ln2 = k (constant), then the differential = k. Hence d(ln2^x)/dx = ln2

How to differentiate ln x 2?

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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