Question

What is the derivative of $\dfrac{\sin x}{1+\cos x}$ ?

Answer 1

In this problem we have to find the derivative of the given $\dfrac{\sin x}{1+\cos x}$. Here we can find the derivative using the division’s derivative formula or the $\dfrac{u}{v}$ method, where $v\ne 0$. We should know that the division’s derivative formula is $g'=\dfrac{u'v-uv'}{{{v}^{2}}}$. We can now further differentiate the terms using some differentiation formula.

Complete step-by-step solution:
Here we have differentiate the given derivative
$\Rightarrow \dfrac{\sin x}{1+\cos x}$
We can now find the derivative using the division’s derivative formula or the $\dfrac{u}{v}$ method, where $v\ne 0$.
We know that the division’s derivative formula is,
$g'=\dfrac{u'v-uv'}{{{v}^{2}}}$. ………. (1)
We can see that from the given data,
$u=\sin x,v=1+\cos x$
We can now substitute the above values in the formula (1), we get
$g'=\dfrac{\left( \sin x \right)'\left( 1+\cos x \right)-\sin x\left( 1+\cos x \right)'}{{{\left( 1+\cos x \right)}^{2}}}$
We can now differentiate the above step, we get
$\Rightarrow g'=\dfrac{\cos x\left( 1+\cos x \right)-\sin x\left( -\sin x \right)}{{{\left( 1+\cos x \right)}^{2}}}$
We can now simplify the above step, we get
$\Rightarrow g'=\dfrac{\cos x+{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\left( 1+\cos x \right)}^{2}}}$
We know that
$\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x=1$
We can now substitute the above formula in the above step, we get
$\Rightarrow g'=\dfrac{\cos x+1}{{{\left( 1+\cos x \right)}^{2}}}$
We can now cancel similar values in the numerator and the denominator, we get
$\Rightarrow g'=\dfrac{1}{\left( 1+\cos x \right)}$
Therefore, the derivative of the given differential equation is $\dfrac{1}{1+\cos x}$.

Note: Students make mistakes while differentiating the equation using derivative using the division’s derivative formula or the $\dfrac{u}{v}$ method, where we differentiate the denominator and keep the numerator as it is and we differentiate the numerator and keep the denominator as it is and we can divide the squared denominator. We should also remember some of the differentiating formulas, which we use in these types of problems.