Question

1-22 Differentiate.

14.\(y = \frac{{\cos x}}{{1 - \sin x}}\)

Short answer

The differentiation of the function \(y = \frac{{\cos x}}{{1 - \sin x}}\) is \(y' = \frac{1}{{1 - \sin x}}\).

Step-by-step solution

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\(y = \frac{{\cos x}}{{1 - \sin x}}\). Differentiate the function w.r.t \(x\) by using the derivatives of trigonometric functions and the quotient rule.

\(\begin{aligned}\frac{{d\left( y \right)}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{{\cos x}}{{1 - \sin x}}} \right)\\ &= \frac{{\left( {1 - \sin x} \right)\frac{d}{{dx}}\left( {\cos x} \right) - \cos x\frac{d}{{dx}}\left( {1 - \sin x} \right)}}{{{{\left( {1 - \sin x} \right)}^2}}}\\ &= \frac{{\left( {1 - \sin x} \right)\left( { - \sin x} \right) - \cos x\left( { - \cos x} \right)}}{{{{\left( {1 - \sin x} \right)}^2}}}\\ &= \frac{{ - \sin x + {{\sin }^2}x + {{\cos }^2}x}}{{{{\left( {1 - \sin x} \right)}^2}}}\end{aligned}\)

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

\(\begin{aligned}y' &= \frac{{ - \sin x + {{\sin }^2}x + {{\cos }^2}x}}{{{{\left( {1 - \sin x} \right)}^2}}}\\ &= \frac{{ - \sin x + 1}}{{{{\left( {1 - \sin x} \right)}^2}}}\\ &= \frac{1}{{1 - \sin x}}\end{aligned}\)

Thus, the derivative of the function \(y = \frac{{\cos x}}{{1 - \sin x}}\) is \(y' = \frac{1}{{1 - \sin x}}\).