differentiate
y=ln(x^5-e^x)
Please show the steps in solving
change it to y=ln u
dy/dx= dy/du * du/dx
and of course, dy/du= 1/u
I will be happy to critique your work.
1/4x * 4/4x = 4/x
Is this correct?
To differentiate the function y = ln(x^5 - e^x), you can use the chain rule.
1. Start by letting u = x^5 - e^x. Then, rewrite the original function as y = ln(u).
2. Calculate the derivative of u with respect to x:
du/dx = d/dx (x^5 - e^x)
= 5x^4 - e^x
3. Now, differentiate y = ln(u) using the chain rule:
dy/dx = (dy/du) * (du/dx)
The derivative of y with respect to u, dy/du, is 1/u.
The derivative of u with respect to x, du/dx, is 5x^4 - e^x.
Therefore, dy/dx = (1/u) * (5x^4 - e^x)
Now, let's simplify the expression obtained:
4. The function y = ln(x^5 - e^x) becomes y = ln(u).
Thus, dy/dx = (1/u) * (5x^4 - e^x).
Finally, to check your answer:
You mentioned that you got 1 / 4x * 4 / 4x = 4/x. Unfortunately, that is not correct.
From step 4, we have dy/dx = (1/u) * (5x^4 - e^x).
To express it more accurately, we can write this as:
dy/dx = (5x^4 - e^x) / (x^5 - e^x).
This is the correct differentiation of the function y = ln(x^5 - e^x).