Question

1-22 Differentiate.

18. \(y = \frac{{\sin t}}{{1 + \tan t}}\)

Short answer

The differentiation of the function \(y = \frac{{\sin t}}{{1 + \tan t}}\) is \(y' = \frac{{\cos t + \sin t - \tan t\sec t}}{{{{\left( {1 + \tan t} \right)}^2}}}\).

Step-by-step solution

Write the formula of the derivatives of trigonometric functions and the quotient rule

\(\begin{aligned}\frac{d}{{dx}}\left( {\tan x} \right) &= {\sec ^2}x\\\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\end{aligned}\)

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

\(\begin{aligned}\frac{{d\left( y \right)}}{{dt}} &= \frac{d}{{dt}}\left( {\frac{{\sin t}}{{1 + \tan t}}} \right)\\ &= \frac{{\left( {1 + \tan t} \right)\frac{d}{{dt}}\left( {\sin t} \right) - \sin t\frac{d}{{dt}}\left( {1 + \tan t} \right)}}{{{{\left( {1 + \tan t} \right)}^2}}}\\ &= \frac{{\left( {1 + \tan t} \right)\left( {\cos t} \right) - \sin t\left( {{{\sec }^2}t} \right)}}{{{{\left( {1 + \tan t} \right)}^2}}}\\ &= \frac{{\cos t + \tan t\cos t - \sin t \cdot \frac{1}{{{{\cos }^2}t}}}}{{{{\left( {1 + \tan t} \right)}^2}}}\\ &= \frac{{\cos t + \frac{{\sin t}}{{\cos t}} \cdot \cos t - \sin t \cdot \frac{1}{{{{\cos }^2}t}}}}{{{{\left( {1 + \tan t} \right)}^2}}}\\ &= \frac{{\cos t + \sin t - \tan t\sec t}}{{{{\left( {1 + \tan t} \right)}^2}}}\end{aligned}\)

Thus, the derivative of the function \(y = \frac{{\sin t}}{{1 + \tan t}}\) is \(y' = \frac{{\cos t + \sin t - \tan t\sec t}}{{{{\left( {1 + \tan t} \right)}^2}}}\).