Question

Find the derivative of the following functions(it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):
$\dfrac{p{{x}^{2}}+qx+r}{ax+b}$

Answer 1

Hint: First, we should know the Leibnitz formula for calculating these type of functions derivative as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{d}{dx}\left( u \right)-u\times \dfrac{d}{dx}\left( v \right)}{{{v}^{2}}}$. Then, by solving the derivative in the equation with the fact that the derivative of x is found by $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. Then, by just simply multiplying the terms in the numerator to get the final result of the expression by using the distributive property as $a\left( c+d \right)=ac+ad$, we get the final result.

Complete step-by-step solution:
In this question, we are supposed to find the derivative of the rational type function.
So, we should know the Leibnitz formula for calculating these type of functions derivative as:
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{d}{dx}\left( u \right)-u\times \dfrac{d}{dx}\left( v \right)}{{{v}^{2}}}$
So, by using the above stated formula in the following given question, we proceed as:
$\dfrac{d}{dx}\left( \dfrac{p{{x}^{2}}+qx+r}{ax+b} \right)=\dfrac{\left( ax+b \right)\times \dfrac{d}{dx}\left( p{{x}^{2}}+qx+r \right)-\left( p{{x}^{2}}+qx+r \right)\times \dfrac{d}{dx}\left( ax+b \right)}{{{\left( ax+b \right)}^{2}}}$
Now, by solving the derivative in the equation with the fact that the derivative of x is found by:
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
So, the above expression gives general idea of the derivative and by using it, we can calculate the individual derivatives in the equation.
Now, by applying the general derivative formula as:
$\dfrac{d}{dx}\left( \dfrac{p{{x}^{2}}+qx+r}{ax+b} \right)=\dfrac{\left( ax+b \right)\times \left( 2px+q \right)-\left( p{{x}^{2}}+qx+r \right)\times \left( a \right)}{{{\left( ax+b \right)}^{2}}}$
Now, by just simply multiplying the terms in the numerator to get the final result of the expression by using the distributive property as:
$a\left( c+d \right)=ac+ad$
So, by following the above statement, we get:
$\dfrac{d}{dx}\left( \dfrac{p{{x}^{2}}+qx+r}{ax+b} \right)=\dfrac{\left( 2ap{{x}^{2}}+aqx+2bpx+bq \right)-\left( ap{{x}^{2}}+aqx+ar \right)}{{{\left( ax+b \right)}^{2}}}$
Now, we need to open the brackets of the numerator and get the final result by subtracting like terms as:
$\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{p{{x}^{2}}+qx+r}{ax+b} \right)=\dfrac{\left( 2ap{{x}^{2}}+aqx+2bpx+bq \right)-ap{{x}^{2}}-aqx-ar}{{{\left( ax+b \right)}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( \dfrac{p{{x}^{2}}+qx+r}{ax+b} \right)=\dfrac{\left( ap{{x}^{2}}+2bpx+bq-ar \right)}{{{\left( ax+b \right)}^{2}}} \\
\end{align}$
So, the derivative of the above given function is $\dfrac{\left( ap{{x}^{2}}+2bpx+bq-ar \right)}{{{\left( ax+b \right)}^{2}}}$.
Hence, the derivative of the function $\dfrac{p{{x}^{2}}+qx+r}{ax+b}$ is $\dfrac{\left( ap{{x}^{2}}+2bpx+bq-ar \right)}{{{\left( ax+b \right)}^{2}}}$.

Note: In this type of question, we should know how to proceed as a derivative of the rational function should not be done separately as we are first doing the derivation of the numerator and then derivation of the denominator and then dividing both the results will not give the exact solution. So, we should keep in mind the rule of the Leibnitz formula to get the value of the differentiation of the function.