what is the derivative of y=ln(x/4)
if y=ln(u)
y' = 1/u u'
y = ln(x/4)
y' = 1/(x/4) * 1/4 = 4/x * 1/4 = 1/x
makes sense, since ln(x/4) = lnx - ln4
so, y' = 1/x
To find the derivative of the function y = ln(x/4), we can use the chain rule. The chain rule states that if we have a composition of functions, f(g(x)), then the derivative is given by the product of the derivative of the outer function (f'(g(x))) and the derivative of the inner function (g'(x)).
In this case, the function y = ln(x/4) can be thought of as the composition of two functions:
f(u) = ln(u), where u = x/4
g(x) = x/4
First, let's find the derivative of the inner function g(x):
g'(x) = d/dx (x/4)
= 1/4
Next, let's find the derivative of the outer function f(u):
f'(u) = d/du (ln(u))
= 1/u
Now, we can apply the chain rule by multiplying the derivatives of the outer and inner functions:
dy/dx = f'(g(x)) * g'(x)
= (1/u) * (1/4)
= 1/(4u)
Since u = x/4, we can replace u with x/4:
dy/dx = 1/(4 * (x/4))
= 1/x
Therefore, the derivative of y = ln(x/4) is dy/dx = 1/x.