I guess I'm lacking in differential equations. I couldn't solve this question. Can you help me?

Rewrite: [tex]t = ln(\frac{2x-1}{x-1})[/tex]
Rewrite [Ln Properties]: [tex]t = ln(2x-1) - ln(x - 1)[/tex]
Differentiate [Ln/Chain Rule/Basic Power Rule]: [tex]\frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1[/tex]
Simplify: [tex]\frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}[/tex]
Rewrite: [tex]\frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}[/tex]
Combine: [tex]\frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}[/tex]
Reciprocate: [tex]\frac{dx}{dt} = -(2x-1)(x-1)[/tex]
Distribute: [tex]\frac{dx}{dt} = (1-2x)(x-1)[/tex]

Supernova Supernova · Answer 2 · 2020-12-25T06:27:34+01:00

Answer:

See below.

Step-by-step explanation:

We are given [tex]\displaystyle ln \Big ( \frac{2x-1}{x-1} \Big ) = t[/tex] and we want to find the first derivative of this function.

We can use the derivative of any function inside a natural log, denoted by [tex]\displaystyle \frac{d}{dx} \text{ln} \ u = \frac{\frac{d}{dx} u}{u}[/tex], where u represents any function.

Let's take the derivative of the whole function with respect to x. This will look like:

[tex]\displaystyle \frac{d}{dx} \displaystyle \Big [ ln \Big ( \frac{2x-1}{x-1} \Big ) = t \Big ] = \frac{\frac{d}{dx} (\frac{2x-1}{x-1}) }{\frac{2x-1}{x-1} } = \frac{dt}{dx}[/tex]

Let's take the derivative of the inside function, [tex]\displaystyle \frac{2x-1}{x-1}[/tex], first. We will need the quotient rule, which is:

[tex]\displaystyle \frac{d}{dx} \Big [ \frac{f(x)}{g(x)} \Big] = \frac{g(x) \cdot \frac{d}{dx}f(x)-f(x)\cdot \frac{d}{dx}g(x) }{[g(x)]^2}[/tex]

Here we have f(x) = 2x - 1 and g(x) = x - 1. Let's plug these values into the formula above:

[tex]\displaystyle \frac{d}{dx} \Big [ \frac{2x-1}{x-1} \Big ] = \frac{[(x-1)\cdot 2 ] - [(2x-1) \cdot 1]}{(x-1)^2}[/tex]
[tex]\displaystyle \frac{d}{dx} \Big [ \frac{2x-1}{x-1} \Big ] = \frac{2x-2-2x+1}{(x-1)^2}[/tex]
[tex]\displaystyle \frac{d}{dx} \Big [ \frac{2x-1}{x-1} \Big ] = \frac{-1}{(x-1)^2}[/tex]

Now, we can substitute this back into the original equation for the derivative of the entire function.

[tex]\displaystyle \frac{dt}{dx} = \frac{\frac{-1}{(x-1)^2} }{\frac{2x-1}{x-1} }[/tex]

Multiply the numerator by the reciprocal of the denominator.

[tex]\displaystyle \frac{dt}{dx} = \frac{-1}{(x-1)^2} \cdot \frac{x-1}{2x-1}[/tex]

The (x - 1)'s cancel out and we are left with:

[tex]\displaystyle \frac{dt}{dx} = \frac{-1}{(x-1)} \cdot \frac{1}{2x-1}[/tex]

This can be further simplified to a single fraction:

[tex]\displaystyle \frac{dt}{dx} =\frac{-1}{(x-1)(2x-1)}[/tex]

Now we have dt/dx, but we want to find dx/dt. Therefore, we can flip the equation and have it in terms of dx/dt:

[tex]\displaystyle \frac{dx}{dt} =\frac{(x-1)(2x-1)}{-1}[/tex]
[tex]\displaystyle \frac{dx}{dt} =-(x-1)(2x-1)[/tex]

This can be further simplified to fit the expression the problem gives for dx/dt:

[tex]\displaystyle \frac{dx}{dt} =(x-1)(1-2x)[/tex]

This is equivalent to the equation in the problem; therefore, the verification is complete.