To find the derivative $f'(x)$ of the function

$f(x) = \frac{bx + c}{kx + a},$

we can apply the quotient rule for derivatives. Recall that if $f(x) = \frac{u(x)}{v(x)}$, then

$f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2}.$

Here, we have:

• $u(x) = bx + c$ with $u'(x) = b$,
• $v(x) = kx + a$ with $v'(x) = k$.

Using the quotient rule, we get:

$f'(x) = \frac{(bx + c)'(kx + a) - (bx + c)(kx + a)'}{(kx + a)^2}.$

Substitute $u'(x) = b$ and $v'(x) = k$:

$f'(x) = \frac{b(kx + a) - (bx + c)k}{(kx + a)^2}.$

Expanding and simplifying the numerator:

$= \frac{bkx + ba - bkx - ck}{(kx + a)^2}.$

Now, notice that $bkx$ terms cancel out:

$= \frac{ba - ck}{(kx + a)^2}.$

Therefore, the derivative is:

$f'(x) = \frac{ba - ck}{(kx + a)^2}.$

Would you like a more detailed breakdown, or do you have any questions?

Here are some related questions for further understanding:

1. How would the answer change if $b = k$?
2. What happens to the derivative if $c = 0$?
3. Can you apply the quotient rule to functions where both numerator and denominator are non-linear?
4. How does the derivative behave as $x \to \infty$?
5. What if $a = 0$? How does it affect $f(x)$ and $f'(x)$?

Tip: When applying the quotient rule, carefully simplify to see if terms cancel out; this often reduces the complexity of your answer.