Question

Differentiate the given trigonometric expression.
$\dfrac{{\sin x - x\cos x}}{{x\sin x + \cos x}}$

Answer 1

Hint: To solve the above question, we will have to use the product rules and quotient rules of differentiation. According to product rule of differentiation, we have:
$\dfrac{d}{{dx}}\left( {f\left( x \right).g\left( x \right)} \right) = g\left( x \right)\left[ {\dfrac{d}{{dx}}f\left( x \right)} \right] + f\left( x \right)\left[ {\dfrac{d}{{dx}}g\left( x \right)} \right]$
The quotient rule of differentiation says that:
$\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\left[ {\dfrac{d}{{dx}}f\left( x \right)} \right] - f\left( x \right)\left[ {\dfrac{d}{{dx}}g\left( x \right)} \right]}}{{{{\left( {g\left( x \right)} \right)}^2}}}$

Complete step-by-step solution -
The function given in question for which we have to find derivative is in the form of $\dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ where $f\left( x \right) = \sin x - x\cos x$ and $g\left( x \right) = xsinx + \cos x$. To differentiate a function of the form $\dfrac{{f\left( x \right)}}{{g\left( x \right)}}$, we will use quotient rule of differentiation says that the derivative of the function $\dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ is calculated as shown below:
$\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\left[ {\dfrac{d}{{dx}}f\left( x \right)} \right] - f\left( x \right)\left[ {\dfrac{d}{{dx}}g\left( x \right)} \right]}}{{{{\left( {g\left( x \right)} \right)}^2}}}$
In our question, we have to find the differentiation of $\dfrac{{\sin x - x\cos x}}{{x\sin x + \cos x}}$. In our case, $f\left( x \right) = \sin x - x\cos x$and $g\left( x \right) = x\sin x + \cos x$. Therefore, after applying quotient rule of differentiation, we get:
$\dfrac{d}{{dx}}\left( {\dfrac{{\sin x - x\cos x}}{{x\sin x + \cos x}}} \right) = \dfrac{{\left( {x\sin x + \cos x} \right)\dfrac{d}{{dx}}\left[ {\sin x - x\cos x} \right] - \left( {\sin x - x\cos x} \right)\left[ {\dfrac{d}{{dx}}\left( {x\sin x + \cos x} \right)} \right]}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}$…………………....(i)
Let $sinx - x\cos x = P$ and $x\sin x + \cos x = q$. Now first we will find the derivative of P with respect to x. Now according to sum rule of differentiation, we have,
\[\dfrac{{dP}}{{dx}} = \dfrac{d}{{dx}}\left( {\sin x - x\cos x} \right) = \dfrac{d}{{dx}}\left( {\sin x} \right) - \dfrac{d}{{dx}}\left( {x\cos x} \right)\] …………….……(ii)
Now, we know that \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]. We will calculate the value of \[\dfrac{d}{{dx}}\left( {x\cos x} \right)\] with the help of quotient rule:
$\dfrac{d}{{dx}}\left( {f\left( x \right).g\left( x \right)} \right) = g\left( x \right).\dfrac{d}{{dx}}f\left( x \right) + f\left( x \right).\dfrac{d}{{dx}}g\left( x \right)$
In our case, $f\left( x \right) = x$ and $g\left( x \right) = \cos x$. So we have:
\[\dfrac{d}{{dx}}\left( {x\cos x} \right) = \cos x.\dfrac{d}{{dx}}\left( x \right) + x.\dfrac{d}{{dx}}\left( {\cos x} \right)\]
Now, because \[\dfrac{{dx}}{{dx}} = 1\,\] and \[\dfrac{d}{{dx}}\cos x = - \sin x\]
\[\dfrac{d}{{dx}}\left( {x\cos x} \right) = \cos x.1 + x\left( { - \sin x} \right)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\cos x} \right) = \cos x - x\sin x\]
Now we will put the respective values into (ii). After doing this:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x - x\cos x} \right) = \cos x - \left( {\cos x - x\sin x} \right)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x - x\cos x} \right) = x\sin x\] ……….……(iii)
Now ,we will find the derivative of q with respect to x.
According to sum rule, we have:
\[\dfrac{{dq}}{{dx}} = \dfrac{d}{{dx}}\left( {x\sin x + \cos x} \right) = \dfrac{d}{{dx}}\left( {x\sin x} \right) + \dfrac{d}{{dx}}\cos x\] ……………….……(iv)
Know we know that \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]. The derivative of $x\sin x$can be calculated with the help of quotient rule:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sin x} \right) = \sin x.\dfrac{d}{{dx}}\left( x \right) + x.\dfrac{d}{{dx}}\left( {\sin x} \right)\]
Now, because \[\dfrac{{dx}}{{dx}} = 1\,\] and \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sin x} \right) = \sin x.1 + x.\cos x\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sin x} \right) = \sin x + x\cos x\]
Now, we will put the respective values into (iv)
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sin x + \cos x} \right) = \sin x + x\cos x - \sin x\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sin x + \cos x} \right) = x\cos x\] ……...……(v)
Now, we will put the values of derivatives from (iii) and (v) into (i)
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{\sin x - x\cos x}}{{x\sin x + \cos x}}} \right] = \dfrac{{\left( {x\sin x + \cos x} \right)\left( {x\sin x} \right) - \left( {\sin x - x\cos x} \right)\left( {x\cos x} \right)}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{P}{q}} \right) = \dfrac{{{x^2}{{\sin }^2}x + x\cos x\sin x - x\cos x\sin x + {x^2}{{\cos }^2}x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{P}{q}} \right) = \dfrac{{{x^2}\left( {{{\sin }^2} + {{\cos }^2}x} \right)}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{P}{q}} \right) = \dfrac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin x - x\cos x}}{{x\sin x + \cos x}}} \right) = {\left( {\dfrac{x}{{x\sin x + \cos x}}} \right)^2}\]

Note: Here we need to keep in mind that the derivative which we have found above is not valid when the denominator is zero. Thus when $x\sin x + \cos x = 0$ the differentiation does not exist.