Question

1-22 Differentiate.

13. \(f\left( \theta \right) = \frac{{\sin \theta }}{{1 + \sin \theta }}\)

Short answer

The differentiation of the function \(f\left( \theta \right) = \frac{{\sin \theta }}{{1 + \sin \theta }}\) is \(f'\left( \theta \right) = \frac{1}{{1 + \cos \theta }}\).

Step-by-step solution

 Step 1: Write the formula of the derivatives of trigonometric functions and the quotient rule

\(\begin{aligned}\frac{d}{{dx}}\left( {\cos x} \right) &= - \sin x\\\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\end{aligned}\)

The Quotient rule: If \(f\) and \(g\) are differentiable functions, then\(\frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right) + f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\).

Find the differentiation of the function

Consider the function \(f\left( \theta \right) = \frac{{\sin \theta }}{{1 + \sin \theta }}\). Differentiate the function w.r.t \(\theta \) by using the derivatives of trigonometric functionsand the quotient rule.

\(\begin{aligned}\frac{{d\left( {f\left( \theta \right)} \right)}}{{d\theta }} &= \frac{d}{{d\theta }}\left( {\frac{{\sin \theta }}{{1 + \cos \theta }}} \right)\\ &= \frac{{\left( {1 + \cos \theta } \right)\frac{d}{{d\theta }}\left( {\sin \theta } \right) - \sin \theta \frac{d}{{d\theta }}\left( {1 + \cos \theta } \right)}}{{{{\left( {1 + \cos \theta } \right)}^2}}}\\ &= \frac{{\left( {1 + \cos \theta } \right)\left( {\cos \theta } \right) - \sin \theta \left( {0 - \sin \theta } \right)}}{{{{\left( {1 + \cos \theta } \right)}^2}}}\\ &= \frac{{\cos \theta + {{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\left( {1 + \cos \theta } \right)}^2}}}\end{aligned}\)

Furthermore, use \({\sin ^2}x + {\cos ^2}x = 1\).

\(\begin{array}{c}f'\left( \theta \right) = \frac{{\cos \theta + {{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\left( {1 + \cos \theta } \right)}^2}}}\\ = \frac{{\cos \theta + 1}}{{{{\left( {1 + \cos \theta } \right)}^2}}}\\ = \frac{1}{{1 + \cos \theta }}\end{array}\)

Thus, the derivative of the function \(f\left( \theta \right) = \frac{{\sin \theta }}{{1 + \sin \theta }}\) is \(f'\left( \theta \right) = \frac{1}{{1 + \cos \theta }}\).