Question

How do you use the quotient rule of differentiation to find the derivative of the $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ .

Answer 1

Hint: In the given problem, we are required to differentiate $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ with respect to x. Since, $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ is a quotient function, so we will have to apply quotient rule of differentiation in the process of differentiating $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ . Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly to solve more such types of questions.

Complete step-by-step answer:
To find derivative of $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ with respect to $x$ , we have to find differentiate $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$with respect to $x$.
So, Derivative of $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ with respect to $x$can be calculated as $\dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ .
Now, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ .
Now, using the quotient rule of differentiation, we know that \[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g\left( x \right)}}} \right) = \left[ {\dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{g^2}\left( x \right)}}} \right] \] .
So, Applying product rule to $\dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right)$, we get,
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right) = \left[ {\dfrac{{\left( {cx + d} \right)a - \left( {ax + b} \right)c}}{{{{\left( {cx + d} \right)}^2}}}} \right] \]
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right) = \left[ {\dfrac{{acx + ad - acx - bc}}{{{{\left( {cx + d} \right)}^2}}}} \right] \]
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right) = \left[ {\dfrac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}} \right] \]
So, the derivative of the function given to us $y = \left( {\dfrac{{ax + b}}{{cx + d}}} \right)$ is \[\dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right) = \left[ {\dfrac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}} \right] \] that is calculated using the quotient rule of differentiation.
ns to ease the process and examine the behaviour of function layer by layer.
So, the correct answer is “\[\dfrac{d}{{dx}}\left( {\dfrac{{ax + b}}{{cx + d}}} \right) = \left[ {\dfrac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}} \right] \] ”.

Note: The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric and algebraic functions must be learned by heart in order to find derivatives of complex functions using quotient rule, product or any other rule of differentiation such as chain rule. The quotient rule of differentiation involves differentiating a quotient function or expression and the chain rule of differentiation involves differentiating a composite by introducing ne