Differentiate the following function with respect to x
a) ln((2x-1/1+x)^1/2)
use your rules of logs to simplify this messy expression
I will assume you meant:
y = ln [ (2x-1)/(1+x) ]^(1/2)
= (1/2)( ln(2x-1) - ln(1+x) )
dy/dx = (1/2)( 2/(2x-1) - 1/(1+x) )
= 1/(2x-1) - 1/(2(1+x))
How do you change y= (1/2)( ln(2x-1) - ln(1+x) )
ln u^n = n * ln u
ln u*v = ln u + ln v
ln u/v = ln u - ln v
Better review the basic properties of logs.
To differentiate the function ln((2x-1)/(1+x)^(1/2)) with respect to x, we can use the chain rule and the logarithmic differentiation rule.
The chain rule states that if we have a composite function y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = f'(g(x)) * g'(x).
In this case, let's define f(u) = ln(u) and g(x) = (2x-1)/(1+x)^(1/2).
First, let's find the derivative of f(u) = ln(u). The derivative of ln(u) with respect to u is 1/u. This means that f'(u) = 1/u.
Next, let's find the derivative of g(x) = (2x-1)/(1+x)^(1/2). To do this, we can use the quotient rule. The quotient rule states that if we have a function y = f(x)/g(x), then the derivative of y with respect to x is given by dy/dx = (g(x)*f'(x) - f(x)*g'(x))/(g(x))^2.
In this case, let f(x) = 2x-1 and g(x) = (1+x)^(1/2).
The derivative of f(x) = 2x-1 with respect to x is simply 2.
To find the derivative of g(x) = (1+x)^(1/2), we can use the chain rule. Let's define h(u) = u^(1/2) and k(x) = 1+x.
The derivative of h(u) = u^(1/2) with respect to u is (1/2)u^(-1/2). This means that h'(u) = (1/2)u^(-1/2).
The derivative of k(x) = 1+x with respect to x is simply 1.
Using the chain rule, the derivative of g(x) = (1+x)^(1/2) with respect to x is g'(x) = h'(k(x))*k'(x) = (1/2)(1+x)^(-1/2)*1 = (1/2)(1+x)^(-1/2).
Now, using the quotient rule, the derivative of ln((2x-1)/(1+x)^(1/2)) with respect to x is:
dy/dx = (g(x)*f'(x) - f(x)*g'(x))/(g(x))^2
= (((2x-1)/(1+x)^(1/2))*2 - ln((2x-1)/(1+x)^(1/2))*(1/2)(1+x)^(-1/2))/((1+x)^(1/2))^2
= (2(2x-1) - ln((2x-1)/(1+x)^(1/2))(1/2)(1+x)^(-1/2))/(1+x)
= (4x - 2 - ln((2x-1)/(1+x)^(1/2))(1+x)^(-1/2))/(1+x)
So, the derivative of ln((2x-1)/(1+x)^(1/2)) with respect to x is:
dy/dx = (4x - 2 - ln((2x-1)/(1+x)^(1/2))(1+x)^(-1/2))/(1+x)