Question

$\frac{ax+b}{cx+d}$
$\left(\text{Differentiate with respect to}x,\text{using first principle}\right)$

Answer 1

Find derivative by first principle method

Given:$f\left(x\right)=\frac{ax+b}{cx+d}$

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\frac{a(x+h)+b}{c(x+h)+d}-\frac{ax+b}{cx+d}}{h}$

$\left[f^{\prime }\right(x)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}]$

$\text{Taking LCM and simplify}$

$={lim}_{h\to 0}\frac{\begin{array}{c}\left[a\right(x+h)+b](cx+d)-\\ (ax+b)\left[c\right(x+h)+d]\end{array}}{h(cx+d)\left[c\right(x+h)+d]}$

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\begin{array}{c}\left(\begin{array}{c}ac{x}^2+adx+achx+\\ ahd+bcx+bd\end{array}\right)-\\ \left(\begin{array}{c}ac{x}^2+achx+\\ adx+bcx+bch+bd\end{array}\right)\end{array}}{h(cx+d)\left[c\right(x+h)+d]}$


$={lim}_{h\to 0}\frac{ahd-bch}{h(cx+d)\left(c\right(x+h)+d)}$

$={lim}_{h\to 0}\frac{h(ad-bc)}{h(cx+d)\left(c\right(x+h)+d)}$

$={lim}_{h\to 0}\frac{(ad-bc)}{(cx+d)\left(c\right(x+h)+d)}$

Now, apply the limit, we get

$f^{\prime }\left(x\right)=\frac{ad-bc}{(cx+d{)}^2}$