Find the first derivative of following function log5 (2x+1/3x-1)
recall that the basic log derivatives apply to logs with base e
then let y = log5 ((2x+1)/(3x-1))
= log5 (2x+1) - log5 (3x-1)
dy/dx = (1/ln5)(2/(2x+1)) - (1/ln5)(3/(3x-1))
= (1/ln5) (2/(2x+1) - 3/(3x-1) )
To find the first derivative of the function log5(2x + 1)/(3x - 1), we can apply the logarithmic and chain rule.
Step 1: Rewrite the function using properties of logarithms.
log5(2x + 1)/(3x - 1) = log5(2x + 1) - log5(3x - 1)
Step 2: Apply the logarithmic rule.
The derivative of log(base a)(f(x)) = f'(x) / (f(x) * ln(a))
So, the derivative of log5(2x + 1) can be found using the chain rule.
Step 3: Find the derivative of log5(2x + 1).
Let u = 2x + 1.
Using the chain rule, the derivative of log5(u) with respect to x is:
d/dx(log5(u)) = 1 / (u * ln(5)) * d/dx(u) = 1 / ((2x + 1) * ln(5)) * d/dx(2x + 1) = 2 / ((2x + 1) * ln(5))
Step 4: Find the derivative of log5(3x - 1).
Let v = 3x - 1.
Using the chain rule, the derivative of log5(v) with respect to x is:
d/dx(log5(v)) = 1 / (v * ln(5)) * d/dx(v) = 1 / ((3x - 1) * ln(5)) * d/dx(3x - 1) = 3 / ((3x - 1) * ln(5))
Step 5: Combine the results.
The first derivative of log5(2x + 1)/(3x - 1) is:
d/dx(log5(2x + 1)/(3x - 1)) = d/dx(log5(2x + 1) - log5(3x - 1))
= 2 / ((2x + 1) * ln(5)) - 3 / ((3x - 1) * ln(5))
Therefore, the first derivative of the given function is 2 / ((2x + 1) * ln(5)) - 3 / ((3x - 1) * ln(5)).