Question

How do you use the quotient rule to find the derivative of $y = \dfrac{{1 + \cos x}}{{1 + \sin x}}?$

Answer 1

In this question, we are going to find the derivative of $y$ by using the quotient rule.
First, we write the given function as $y = \dfrac{u}{v}$ and then apply the quotient rule to the given expression.
Differentiate wherever necessary and use trigonometric identities to simplify the derivative.
Hence we can get the required result.

Formula used: The quotient rule states that
$\dfrac{{dy}}{{dx}} = \dfrac{{vu' - uv'}}{{{v^2}}}$
$\dfrac{{dy}}{{dx}} =$ Derivative of $y$ with respect to $x$
$v =$ Variable $v$
$u =$ Variable $u$
$v' =$ Derivative of $v$ with respect to $x$
$u' =$ Derivative of $u$ with respect to $x$

Complete step-by-step solution:
In this question, we are going to derive the value of $y$ by using the quotient rule.
The given function is of the form $y = \dfrac{u}{v}$
By assigning $u$ and $v$ equal to the numerator and denominator respectively.
Now by applying the quotient rule to the given function we get,
$\dfrac{{dy}}{{dx}} = \dfrac{{vu' - uv'}}{{{v^2}}}$
Here $u = 1 + \cos x,v = 1 + \sin x$
$u' = - \sin x,\,v' = \cos x$
From the identities of trigonometric function derivatives, we know that the derivative of sine $x$ is cosine $x$ and derivative of cosine $x$ is minus sine $x$.
This could be proven using Euler’s formula, but for our purposes we shall accept these without proof.
The derivative of any constant is zero, and the derivative of a sum is equal to the sum of the derivatives.
Substituting the above values in the quotient rule we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{1 + \sin x\left( { - \sin x} \right) - \left( {1 + \cos x} \right)\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}}$
Multiplying the terms inside we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin x - \left( {{{\sin }^2}x} \right) - \left( {\cos x + {{\cos }^2}x} \right)}}{{{{\left( {1 + \sin x} \right)}^2}}}$
Rewrite the above term as
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin x - {{\sin }^2}x - \cos x - {{\cos }^2}x}}{{{{\left( {1 + \sin x} \right)}^2}}}$
Combining the like terms together we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - (\sin x + \cos x) - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\left( {1 + \sin x} \right)}^2}}}$
Since ${\sin ^2}x + {\cos ^2}x = 1$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - (\sin x + \cos x) - 1}}{{{{\left( {1 + \sin x} \right)}^2}}}$
On rewriting we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin x - \cos x - 1}}{{{{\left( {1 + \sin x} \right)}^2}}}$
Then we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sin x + \cos x + 1}}{{{{\left( {1 + \sin x} \right)}^2}}}$
That is $y' = - \dfrac{{1 + \cos x + \sin x}}{{{{\left( {1 + \sin x} \right)}^2}}}$

Hence the derivative of $y$ is $- \dfrac{{1 + \cos x + \sin x}}{{{{\left( {1 + \sin x} \right)}^2}}}$

Note: When using the quotient rule, it is important to designate our functions $u$ and $v$ in such a way to make things simple while still being accurate. In the above case, declaring $\cos x$ as $u$ would not allow using the quotient rule efficiently, as our numerator then would be $1 + u$.