Hi! I need help finding the derivative of this function.
f(x)= ln [(x(x-1))/(x-2)]
Thank you.
f(x)= ln[(x(x-1))/(x-2)]
f(x)= ln(x(x-1)-ln(x-2)
f(x)= ln(x)+ln(x-1)-ln(x-2)
f'(x)=1/x+1/(x-1)+1/(x-2)
Thanks Sarah!
Of course! To find the derivative of the function f(x) = ln [(x(x-1))/(x-2)], we can use the rules of differentiation.
First, let's simplify the function before taking the derivative. Using the properties of logarithms, we can rewrite the function as:
f(x) = ln[x(x-1)] - ln(x-2)
Now, we can find the derivative of each term separately using the rules of differentiation.
For the first term, ln[x(x-1)], we use the product rule. Let u(x) = x and v(x) = (x-1). Therefore, we have:
u'(x) = 1 (the derivative of x is 1)
v'(x) = 1 (the derivative of x-1 is 1)
Now using the product rule (d/dx)[u(x)*v(x)] = u'(x)*v(x) + u(x)*v'(x), we get:
[first term]'(x) = (1*(x-1)) + (x*1) = x - 1 + x = 2x - 1
For the second term, ln(x-2), we use the chain rule. Let u(x) = x-2. Therefore, we have:
u'(x) = 1 (the derivative of x-2 is 1)
Now using the chain rule (d/dx)[ln(u(x))] = u'(x)/u(x), we get:
[second term]'(x) = 1/(x-2)
Therefore, the derivative of the function f(x) = ln [(x(x-1))/(x-2)] is:
f'(x) = [first term]'(x) - [second term]'(x)
= 2x - 1 - 1/(x-2)
And that's the derivative of the given function.