Question

1-22 Differentiate.

16.\(f\left( t \right) = \frac{{\cot \,t}}{{{e^t}}}\)

Short answer

The differentiation of the function \(f\left( t \right) = \frac{{\cot t}}{{{e^t}}}\) is \(f'\left( t \right) = \frac{{ - {{\csc }^2}t - \cot t}}{{{e^t}}}\).

Step-by-step solution

Write the formula of the derivatives of trigonometric functions, exponential function and the quotient rule

\(\begin{aligned}\frac{d}{{dx}}\left( {\cot } \right) &= {\csc ^2}x\\\frac{d}{{dx}}\left( {{e^x}} \right) &= {e^x}\end{aligned}\)

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\(f\left( t \right) = \frac{{\cot t}}{{{e^t}}}\). Differentiate the function w.r.t \(t\) by using the derivatives of trigonometric functions, exponential function and the quotient rule.

\(\begin{aligned}\frac{{d\left( {f\left( t \right)} \right)}}{{dt}} &= \frac{d}{{dt}}\left( {\frac{{\cot t}}{{{e^t}}}} \right)\\ &= \frac{{{e^t}\frac{d}{{dt}}\left( {\cot t} \right) - \cot t\frac{d}{{dt}}\left( {{e^t}} \right)}}{{{{\left( {{e^t}} \right)}^2}}}\\ &= \frac{{{e^t}\left( { - {{\csc }^2}t} \right) - \cot t\left( {{e^t}} \right)}}{{{{\left( {{e^t}} \right)}^2}}}\\ &= \frac{{{e^t}\left( { - {{\csc }^2}t - \cot t} \right)}}{{{{\left( {{e^t}} \right)}^2}}}\\ &= \frac{{ - {{\csc }^2}t - \cot t}}{{{e^t}}}\end{aligned}\)

Thus, the derivative of the function \(f\left( t \right) = \frac{{\cot t}}{{{e^t}}}\) is \(f'\left( t \right) = \frac{{ - {{\csc }^2}t - \cot t}}{{{e^t}}}\).