Question

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

Answer 1

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

• Equality Properties

• Reciprocals

Algebra II

• Log/Ln Property: $ln(\frac{a}{b} ) = ln(a) - ln(b)$

Calculus

Derivatives

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative of Ln: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

$ln(\frac{2x-1}{x-1} )=t$

Step 2: Differentiate

1. Rewrite:                                                                                                         $t = ln(\frac{2x-1}{x-1})$
2. Rewrite [Ln Properties]:                                                                                 $t = ln(2x-1) - ln(x - 1)$
3. Differentiate [Ln/Chain Rule/Basic Power Rule]:                                         $\frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1$
4. Simplify:                                                                                                          $\frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}$
5. Rewrite:                                                                                                          $\frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}$
6. Combine:                                                                                                       $\frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}$
7. Reciprocate:                                                                                                  $\frac{dx}{dt} = -(2x-1)(x-1)$
8. Distribute:                                                                                                         $\frac{dx}{dt} = (1-2x)(x-1)$

Answer 2

Answer:

See below.

Step-by-step explanation:

We are given $\displaystyle ln \Big ( \frac{2x-1}{x-1} \Big ) = t$ 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 $\displaystyle \frac{d}{dx} \text{ln} \ u = \frac{\frac{d}{dx} u}{u}$, where u represents any function.

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

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

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

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

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

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

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

• $\displaystyle \frac{dt}{dx} = \frac{\frac{-1}{(x-1)^2} }{\frac{2x-1}{x-1} }$  

Multiply the numerator by the reciprocal of the denominator.

• $\displaystyle \frac{dt}{dx} = \frac{-1}{(x-1)^2} \cdot \frac{x-1}{2x-1}$

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

• $\displaystyle \frac{dt}{dx} = \frac{-1}{(x-1)} \cdot \frac{1}{2x-1}$

This can be further simplified to a single fraction:

• $\displaystyle \frac{dt}{dx} =\frac{-1}{(x-1)(2x-1)}$  

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:

• $\displaystyle \frac{dx}{dt} =\frac{(x-1)(2x-1)}{-1}$
• $\displaystyle \frac{dx}{dt} =-(x-1)(2x-1)$

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

• $\displaystyle \frac{dx}{dt} =(x-1)(1-2x)$

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