Question

What is the derivative of ln e^x / e^x -1

Answer 1

The derivative of ln(e^x) / (e^x - 1) is found by simplifying ln(e^x) to x and applying the quotient rule, which results in -xe^x - 1 / (e^x - 1)^2.

Step-by-step explanation:

The student is asking for the derivative of the function ln(e^x) / (e^x - 1). To find the derivative, we need to apply the rules of differentiation. To address the logarithm part, we use the fact that the natural logarithm (ln) and the exponential function (e^x) are inverse functions; hence, ln(e^x) simplifies to x. However, we must also apply the quotient rule to differentiate the given expression as a whole.

Step 1: Simplify the expression

ln(e^x) simplifies to x, so the function becomes x / (e^x - 1).

Step 2: Apply the Quotient Rule

The quotient rule states that if we have a function h(x) = f(x)/g(x), then its derivative h'(x) = (f'(x)g(x) - f(x)g'(x)) / [g(x)]^2.

Applying this to x / (e^x - 1), we get:

Derivative: (1*(e^x - 1) - x*e^x) / (e^x - 1)^2

Simplify the numerator: e^x - 1 - x*e^x = -xe^x - 1

So, the final derivative is -xe^x - 1 / (e^x - 1)^2.

Answer 2

I assume you mean


`(\mathrm d)/(\mathrm dx)\ln(e^x)/(e^x-1)`

The chain rule says this is equivalent to


`((\mathrm d)/(\mathrm dx)(e^x)/(e^x-1))/((e^x)/(e^x-1))`

The numerator can be worked out via the quotient rule:


`(\mathrm d)/(\mathrm dx)(e^x)/(e^x-1)=(e^x(e^x-1)-e^x* e^x)/((e^x-1)^2)=-(e^x)/((e^x-1)^2)`

We're left with


`(-(e^x)/((e^x-1)^2))/((e^x)/(e^x-1))=-\frac1{e^x-1}=\frac1{1-e^x}`