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Question: How do you differentiate \( f\left( x \right) = \dfrac{{ax + b}}{{cx + d}} \)...

How do you differentiate f(x)=ax+bcx+df\left( x \right) = \dfrac{{ax + b}}{{cx + d}}

Explanation

Solution

Hint : In the question, we have to find the derivative of the function of xx . For that applying the formula for derivative which is g(x)f′(x)−f(x)g′(x)(g(x))2\dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{(g\left( x \right))}^2}}} where the f(x)g(x)\dfrac{{f\left( x \right)}}{{g\left( x \right)}} represents the division of the function.

Complete step-by-step answer :
In this question, we have to differentiate this function.
The given function is f(x)=ax+bcx+df\left( x \right) = \dfrac{{ax + b}}{{cx + d}} in this expression xx is a variable and we have to differentiate it with respect to xx
Applying the formula for differentiation which states that to find the derivative of f(x)f\left( x \right) divided by g(x)g\left( x \right) . Take the product of g(x)g\left( x \right) times the derivative of f(x)f\left( x \right) subtracted by product of f(x)f\left( x \right) and the derivative of g(x)g\left( x \right) divided by the square of the g(x)g\left( x \right)
Let the function be f(x)g(x)\dfrac{{f\left( x \right)}}{{g\left( x \right)}} then the formula of differentiation of this expression is
g(x)f′(x)−f(x)g′(x)(g(x))2\dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{(g\left( x \right))}^2}}} here g′(x)g'\left( x \right) represents the derivative of g(x)g\left( x \right) and f(x)f\left( x \right) represents the derivative of f′(x)f'\left( x \right)
Now we are differentiating the given expression,
f′(x)=(cx+d)d(ax+b)dx−(ax+b)d(cx+d)dx(cx+d)2f'\left( x \right) = \dfrac{{\left( {cx + d} \right)\dfrac{{d\left( {ax + b} \right)}}{{dx}} - \left( {ax + b} \right)\dfrac{{d\left( {cx + d} \right)}}{{dx}}}}{{{{\left( {cx + d} \right)}^2}}}
Solving,
⇒f′(x)=(cx+d)a−(ax+b)c(cx+d)2\Rightarrow f'\left( x \right) = \dfrac{{\left( {cx + d} \right)a - \left( {ax + b} \right)c}}{{{{\left( {cx + d} \right)}^2}}}
⇒f′(x)=cxa+da−axc−bc(cx+d)2\Rightarrow f'\left( x \right) = \dfrac{{cxa + da - axc - bc}}{{{{\left( {cx + d} \right)}^2}}}
Hence, the required answer is
⇒f′(x)=da−bc(cx+d)2\Rightarrow f'\left( x \right) = \dfrac{{da - bc}}{{{{\left( {cx + d} \right)}^2}}}
So, the correct answer is “ f′(x)=da−bc(cx+d)2f'\left( x \right) = \dfrac{{da - bc}}{{{{\left( {cx + d} \right)}^2}}} ”.

Note : Be careful while taking the derivative because students get confused in the formula as to which function should come first. Then the denominator function should be first taken and then take the derivative of the numerator and then the numerator into the derivative of the denominator.