Find f'(x) if f(x)= Logx(x^2-5x+6)
NOTE:
that logx is a sub x
let y = logx (x^2 - 5x + 6)
x^y = x^2 - 5x + 6
ln both sides
ln (x^y) = ln(x^2 - 5x + 6)
y lnx = ln(x^2 - 5x + 6)
differentiate implicityly
y(1/x) + (lnx) dy/dx = (2x-5)/(x^2-5x+6)
dy/dx = ((2x-5)/(x^2-5x+6) - y/x)/lnx
To find the derivative of f(x), which is denoted as f'(x), we can use the logarithmic differentiation technique. Here's the step-by-step process:
1. Start by expressing the function f(x) as a natural logarithm: f(x) = ln(x^2 - 5x + 6).
2. Take the natural logarithm of both sides of the equation: ln(f(x)) = ln(ln(x^2 - 5x + 6)).
3. Differentiate both sides of the equation using the chain rule:
On the left-hand side, we have d/dx [ln(f(x))] = f'(x)/f(x).
On the right-hand side, since we have a composition of functions, we will apply the chain rule:
d/dx [ln(ln(x^2 - 5x + 6))] = 1 / ln(x^2 - 5x + 6) * d/dx (ln(x^2 - 5x + 6)).
4. Now, let's calculate the derivative of ln(x^2 - 5x + 6). Using the chain rule, we have:
d/dx [ln(x^2 - 5x + 6)] = 1 / (x^2 - 5x + 6) * d/dx (x^2 - 5x + 6).
5. Differentiating the expression x^2 - 5x + 6 gives us 2x - 5.
6. Plugging the result back into the previous step, we have:
d/dx [ln(x^2 - 5x + 6)] = 1 / (x^2 - 5x + 6) * (2x - 5).
7. Substituting the result from step 6 back into step 3, we get:
f'(x)/f(x) = 1 / ln(x^2 - 5x + 6) * (2x - 5).
8. Solve for f'(x) by multiplying both sides by f(x):
f'(x) = f(x) * [1 / ln(x^2 - 5x + 6) * (2x - 5)].
9. Finally, substitute back f(x) = ln(x^2 - 5x + 6) into the equation from step 8:
f'(x) = ln(x^2 - 5x + 6) * [1 / ln(x^2 - 5x + 6) * (2x - 5)].
Simplifying further, we get:
f'(x) = 2x - 5.
Thus, the derivative of f(x) = logx(x^2 - 5x + 6) is f'(x) = 2x - 5.