Question

Find dy/dx if y=ln(x^2/e^x)

Answer 1

The derivative of the expression
`\text{y} = \ln((x^2)/(e^x))` is
`(dy)/(dx) = (2 - x)/(x)`

How to determine the derivative of the expression

From the question, we have the following parameters that can be used in our computation:

`\text{y} = \ln((x^2)/(e^x))`

Let

`u = (x^2)/(e^x)`

So, we have

y = ln(u)

Differentiating u, we make use of the quotient rule which states that

if y = u/v, then

`(dy)/(dx) = (vu' - uv')/(v^2)`

So, we have

`(du)/(dx) = (e^x * 2x - x^2 * e^x)/(e^(2x))`

Expand and evaluate

`(du)/(dx) = (2x - x^2)/(e^(x))`

Recall that

y = ln(u)

So:

`(dy)/(du) = (1)/(u)`

This means that

`(dy)/(du) = (e^x)/(x^2)`

The derivative of the expression is then calculated as

`(dy)/(dx) = (dy)/(du) * (du)/(dx)`

Substitute the known values into the equation

`(dy)/(dx) = (e^x)/(x^2) * (2x - x^2)/(e^(x))`

Evaluate the product

`(dy)/(dx) = (2x - x^2)/(x^(2))`

Simplify

`(dy)/(dx) = (2 - x)/(x)`

Hence, the derivative is (2 - x)/x

Answer 2

To find the derivative dy/dx of y=ln(x^2/e^x), rewrite the expression using logarithm properties and then differentiate each part separately. Apply the chain rule where necessary. The final derivative is dy/dx = 2/x - 1.

Step-by-step explanation:

To find the derivative of dy/dx for the function y=ln(x^2/e^x), we will apply the chain rule and the properties of logarithms. First, rewrite the function using the logarithm property that allows division inside a logarithm to be expressed as a subtraction outside of the logarithm:

y = ln(x^2) - ln(e^x)

Now, take the derivative of each term separately. Remember that the derivative of ln(u) is 1/u * du/dx:

• For the first term, the derivative is 2/x.
• For the second term, the derivative is 1/e^x multiplied by the derivative of e^x, which is e^x itself, simplifying to 1.

The final derivative dy/dx is then:

dy/dx = 2/x - 1