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How do you find the derivative of $\dfrac{\sin x}{1+\cos x}$?

seo-qna
Last updated date: 20th Jun 2024
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Answer
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Hint: In this problem we need to calculate the derivative of the given function. We can observe that the given function is a fraction with numerator $\sin x$, denominator $1+\cos x$. In differentiation we have the formula $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{{{u}^{'}}v-u{{v}^{'}}}{{{v}^{2}}}$. So, we will compare the given function with $\dfrac{u}{v}$ and write the values of $u$, $v$. After knowing the values of $u$, $v$ we will differentiate the both the values to get the values of ${{u}^{'}}$, ${{v}^{'}}$. After knowing the value of ${{u}^{'}}$, ${{v}^{'}}$ we will use the division’s derivative formula and simplify the equation to get the required result.

Complete step by step solution:
Given that, $\dfrac{\sin x}{1+\cos x}$.
Comparing the above fraction with $\dfrac{u}{v}$, then we will get
$u=\sin x$, $v=1+\cos x$.
Considering the value $u=\sin x$.
Differentiating the above equation with respect to $x$, then we will get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( \sin x \right)$
We have the differentiation formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, then we will have
$\Rightarrow {{u}^{'}}=\cos x$
Considering the value $v=1+\cos x$.
Differentiating the above equation with respect to $x$, then we will get
$\Rightarrow \dfrac{dv}{dx}=\dfrac{d}{dx}\left( 1+\cos x \right)$
We know that the differentiation of constant is zero and $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, then we will have
$\Rightarrow {{v}^{'}}=-\sin x$
Now the derivative of the given fraction is given by
$\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{{{u}^{'}}v-u{{v}^{'}}}{{{v}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{\cos x\left( 1+\cos x \right)-\sin x\left( -\sin x \right)}{{{\left( 1+\cos x \right)}^{2}}} \\
\end{align}$
Simplifying the above equation, then we will have
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{\cos x+{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\left( 1+\cos x \right)}^{2}}}$
We have the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, then the above equation is modified as
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{1+\cos x}{{{\left( 1+\cos x \right)}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{1}{1+\cos x} \\
\end{align}$

Hence the derivative of the given equation $\dfrac{\sin x}{1+\cos x}$ is $\dfrac{1}{1+\cos x}$.

Note: In this problem we can observe that ${{v}^{'}}=-\sin x=-u$. For this type of equation calculation of the integration is also simple. We have the integration formula $\int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}dx}=\log \left| f\left( x \right) \right|+C$. We will use this formula and simplify it to get the integration value.