what are the steps and explanations for differantiating this function?
y = ln(e^-x + xe^-x)
replace the <> function with u.
y=ln u
y'=1/u du/dx
where du/dx is d/dx (e^-x(1+ x))
to that as a d(vw)/dx
Then put it all together.
thanks, but I have another question.
How does ln(e^-x) = -x
I know that the derivative of ln is 1/x so it would be 1/(e^-x) but how does that simplify to -x? I am confused....
ln of e^a= a by definition.
What does log mean if not that?
To differentiate the function y = ln(e^-x + xe^-x), we can follow these steps:
Step 1: Identify the function parts
In this case, we have a composition of logarithmic and exponential functions. To simplify this, let's assign a variable u to the expression e^-x + xe^-x. So we have y = ln(u).
Step 2: Use the chain rule
The chain rule is a basic rule in calculus that allows us to differentiate composite functions. For a function h(g(x)), the chain rule states that the derivative is given by h'(g(x)) * g'(x).
Applying the chain rule to our function y = ln(u), we have dy/dx = dy/du * du/dx.
Step 3: Find the derivatives
To find dy/du, we differentiate ln(u) with respect to u. The derivative of ln(u) with respect to u is 1/u. So, dy/du = 1/u.
Now, let's find du/dx. We have u = e^-x + xe^-x. First, let's differentiate e^-x. The derivative of e^x is simply e^x, so the derivative of e^-x is -e^-x.
For the second term, let's treat xe^-x as a product of two functions: x and e^-x. Using the product rule, we have (x)' * (e^-x) + (x) * (e^-x)', which simplifies to e^-x - xe^-x.
Therefore, du/dx = -e^-x + e^-x - xe^-x = -xe^-x.
Step 4: Combine the derivatives
Now that we have dy/du = 1/u and du/dx = -xe^-x, we can substitute these results back into the chain rule equation: dy/dx = dy/du * du/dx.
Substituting, we have dy/dx = (1/u) * (-xe^-x) = -xe^-x/u.
Step 5: Substitute back u
Remember that u represents the expression e^-x + xe^-x. Therefore, substituting back, we have dy/dx = -xe^-x/(e^-x + xe^-x).
And that's how you differentiate the function y = ln(e^-x + xe^-x).