Differentiate the following w.r.t. as indicated :
`(ax+b)/(cx+d)" w.r.t. "(a'x+b')/(c'x+d')`

Answer

To differentiate the function \( \frac{ax+b}{cx+d} \) with respect to \( \frac{a'x+b'}{c'x+d'} \), we can follow these steps:

### Step 1: Define the Functions
Let:
- \( u = \frac{ax+b}{cx+d} \)
- \( v = \frac{a'x+b'}{c'x+d'} \)

### Step 2: Use the Chain Rule
We need to find \( \frac{du}{dv} \). By the chain rule, we can express this as:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]
where \( x \) is the variable we will differentiate with respect to.

### Step 3: Differentiate \( u \) with Respect to \( x \)
Using the quotient rule for differentiation:
\[ \frac{du}{dx} = \frac{(cx+d)(a) - (ax+b)(c)}{(cx+d)^2} \]
Simplifying this:
\[ \frac{du}{dx} = \frac{acx + ad - acx - bc}{(cx+d)^2} = \frac{ad - bc}{(cx+d)^2} \]

### Step 4: Differentiate \( v \) with Respect to \( x \)
Similarly, we differentiate \( v \):
\[ \frac{dv}{dx} = \frac{(c'x+d')(a') - (a'x+b')(c')}{(c'x+d')^2} \]
Simplifying this:
\[ \frac{dv}{dx} = \frac{a'c'x + a'd' - a'c'x - b'c'}{(c'x+d')^2} = \frac{a'd' - b'c'}{(c'x+d')^2} \]

### Step 5: Substitute into the Chain Rule
Now we can substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into the chain rule expression:
\[ \frac{du}{dv} = \frac{\frac{ad - bc}{(cx+d)^2}}{\frac{a'd' - b'c'}{(c'x+d')^2}} = \frac{(ad - bc)(c'x + d')^2}{(a'd' - b'c')(cx + d)^2} \]

### Final Result
Thus, the derivative of \( \frac{ax+b}{cx+d} \) with respect to \( \frac{a'x+b'}{c'x+d'} \) is:
\[ \frac{du}{dv} = \frac{(ad - bc)(c'x + d')^2}{(a'd' - b'c')(cx + d)^2} \]