Find the derivative of y = ln |cos(ln)|.
I got -tan(ln x)/x, is this correct?
Is that a typo ?
Should it say: y = ln |cos(ln x) | ?
If so, then you are correct, good job.
Ahh yes that was a typo! Thank you!!
To find the derivative of y = ln |cos(ln x)|, we can use the chain rule and the derivative of the natural logarithm function.
Let's break down the process step by step:
Step 1: First, let's rewrite the function using the absolute value and the chain rule.
y = ln |cos(ln x)|
Step 2: Take the natural logarithm of both sides, which cancels out the ln function:
e^y = |cos(ln x)|
Step 3: Now, let's differentiate both sides using the chain rule.
d/dx(e^y) = d/dx(|cos(ln x)|)
Step 4: Recall that the derivative of e^y with respect to x is e^y times dy/dx (using the chain rule).
e^y * dy/dx = d/dx(|cos(ln x)|)
Step 5: Now, let's differentiate the absolute value function and the composition inside the absolute value using the chain rule.
e^y * dy/dx = d/dx(cos(ln x)) for x > 0, and -d/dx(cos(ln x)) for x < 0
Step 6: Our function contains the natural logarithm function, so we need to use the chain rule again.
e^y * dy/dx = -d/dx(cos(ln x))
Step 7: The derivative of cos(ln x) with respect to x can be found by applying the chain rule once more.
e^y * dy/dx = sin(ln x) * d/dx(ln x)
Step 8: Use the chain rule for d/dx(ln x) by considering it as ln(u) where u = x.
e^y * dy/dx = sin(ln x) * 1/x
Step 9: Finally, solve for dy/dx by dividing both sides by e^y.
dy/dx = (sin(ln x) * 1/x) / e^y
Step 10: To simplify the expression, recall that e^y is equal to |cos(ln x)|, so we substitute it back in.
dy/dx = (sin(ln x) * 1/x) / |cos(ln x)|
So, the correct derivative of y = ln |cos(ln x)| is dy/dx = (sin(ln x) * 1/x) / |cos(ln x)|.
It seems that your result, -tan(ln x)/x, is incorrect.