What is the DERIVATIVE of:
ln x^2 * ln 6 -(product rule)
you are told to apply the product rule, so apply it
(fg)' = f'g + fg'
f = x^2
g = ln x
(I assume ln 6) is a typo.
Make the substitution u = x^2 and then use the product rule
d/dx (u) = d/dx(lnu)*du/dx
= 1/u*(du/dx)
= (1/x^2)(2x)
= 2/x
Don't forget the ln6 factor
oops - I see I missed the ln(x^2)
drwls caught it.
However, I still think ln6 is a typo, or there's no need for the product rule; just the chain rule.
So, if ln6 should be lnx then
f = ln x^2
g = ln x
note that f = 2 lnx
so, fg = 2ln^2(x)
but that is again moving away from the product rule. Good to check, though:
(2 ln^2 x)' = 2 * 2lnx * 1/x = 4/x ln x
the way it stands, there is no need for the product rule, since the ln6 is merely a constant
so
y = ln x^2 * ln 6
= 2lnx (ln6)
= 2ln6 (lnx)
dy/dx = 2ln6 (1/x)
To find the derivative of the product of two functions, such as ln(x^2) and ln(6), we use the product rule.
The product rule states that if we have two functions, f(x) and g(x), then the derivative of their product, f(x) * g(x), is given by the formula:
(f(x) * g(x))' = f'(x) * g(x) + f(x) * g'(x)
Now, let's apply the product rule to find the derivative of ln(x^2) * ln(6).
We start by identifying f(x) and g(x) as follows:
f(x) = ln(x^2)
g(x) = ln(6)
Next, we find the derivatives of f(x) and g(x).
The derivative of ln(x^2) can be found using the chain rule and is given by:
f'(x) = (2/x^2) * ln(x^2)
The derivative of ln(6) with respect to x is zero since ln(6) is a constant.
Now, we can substitute our values into the product rule formula:
(f(x) * g(x))' = f'(x) * g(x) + f(x) * g'(x)
(ln(x^2) * ln(6))' = (2/x^2) * ln(x^2) * ln(6) + ln(x^2) * 0
Simplifying further:
(ln(x^2) * ln(6))' = (2/x^2) * ln(x^2) * ln(6)
Therefore, the derivative of ln(x^2) * ln(6) with respect to x is (2/x^2) * ln(x^2) * ln(6).