Question

Find the derivative of the following functions ( it is to be understand that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers ): \[\left( px+q \right)\left( \dfrac{r}{x}+s \right)\]

Answer 1

Hint: First of all, let us assume the given function as f(x). Now we will write coefficients of x at one side, coefficient of \[\left( \dfrac{1}{x} \right)\] at one side and constants at other side. Let us assume this equation as equation (1). Now we will differentiate equation (1) with x. Let us assume this function as\[{f}'(x)\].
This will give the derivative of the given function.

Complete step by step answer:
Before solving the problem, let us assume the given function \[\left( px+q \right)\left( \dfrac{r}{x}+s \right)\] as f(x) where a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers.
 \[\Rightarrow f(x)=\left( px+q \right)\left( \dfrac{r}{x}+s \right)\]
Now by multiplication on R.H.S, we get
\[\begin{align}
  & \Rightarrow f(x)=(px+q)\left( \dfrac{r}{x} \right)+\left( px+q \right)(s) \\
 & \Rightarrow f(x)=pr+\dfrac{qr}{x}+psx+qs \\
 & \Rightarrow f(x)=psx+\dfrac{qr}{x}+\left( pr+qs \right)...(1) \\
\end{align}\]
Now we will differentiate equation (1) with x on both sides.
We also know that the differentiation of constant is equal to zero.
 \[\begin{align}
  & \Rightarrow {f}'(x)=\dfrac{d}{dx}\left( psx+\dfrac{qr}{x}+\left( pr+qs \right) \right) \\
 & \Rightarrow {f}'(x)=\dfrac{d}{dx}(psx)+\dfrac{d}{dx}\left( \dfrac{qr}{x} \right)+\dfrac{d}{dx}(pr+qs) \\
 & \Rightarrow {f}'(x)=ps\dfrac{d}{dx}(x)+(qr)\dfrac{d}{dx}\left( \dfrac{1}{x} \right) \\
\end{align}\]
We know that \[\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}\]
\[\begin{align}
  & \Rightarrow {f}'(x)=ps(1)+(qr)\left( \dfrac{-1}{{{x}^{2}}} \right) \\
 & \Rightarrow {f}'(x)=ps-\dfrac{qr}{{{x}^{2}}}....(2) \\
 & \\
\end{align}\]
So, the derivative of \[\left( px+q \right)\left( \dfrac{r}{x}+s \right)\] is equal to \[ps-\dfrac{qr}{{{x}^{2}}}\].

Note:
This problem can be solved in an alternative method. Let us assume f(x) as \[\left( px+q \right)\left( \dfrac{r}{x}+s \right)\]. Now we have to find derivative of f(x). Let the derivative of f(x) as \[{f}'(x)\].
\[\Rightarrow f(x)=\left( px+q \right)\left( \dfrac{r}{x}+s \right)\]
Let us assume \[\left( px+q \right)\] as u and \[\left( \dfrac{r}{x}+s \right)\] as v. We know that \[\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
So, we get
\[\begin{align}
  & \Rightarrow {{f}^{|}}(x)=\left( \dfrac{r}{x}+s \right)\dfrac{d}{dx}(px+q)+\left( px+q \right)\dfrac{d}{dx}\left( \dfrac{r}{x}+s \right) \\
 & \Rightarrow {{f}^{|}}(x)=\left( \dfrac{r}{x}+s \right)(p)+(px+q)\left( \dfrac{-r}{{{x}^{2}}} \right) \\
 & \Rightarrow {{f}^{|}}(x)=ps-\dfrac{qr}{{{x}^{2}}} \\
\end{align}\]
Hence, the derivative of \[\left( px+q \right)\left( \dfrac{r}{x}+s \right)\] is equal to \[ps-\dfrac{qr}{{{x}^{2}}}\].