Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Compute the derivatives of a. \(\quad y=2 \sin t \cos t\) b. \(y=\sin ^{2} t+\cos ^{2} t\) c. \(\quad y=\sec t=\frac{1}{\cos t}\) d. \(y=\cot t\) e. \(y=\ln \cos t\) f. \(y=\sin ^{2} t-\cos ^{2} t\) g. \(\quad y=\csc t=\frac{1}{\sin t}\) h. \(y=\sec ^{2} t\) i. \(\quad y=e^{\cos t}\) j. \(\quad y=\ln (\sec t)\) k. \(\quad y=e^{-t} \sin t\) l. \(y=\tan ^{2} t\) m. \(\quad y=\frac{e^{2 t}}{400}\) n. \(y=\ln \left(\cos ^{20} t\right)\) o. \(\quad y=\frac{\ln t^{2}}{30}\)

Step-by-Step Solution

Answer

The derivatives are: (a) \(2\cos 2t\), (b) 0, (c) \(\sec t \tan t\), (d) \(-\csc^2 t\), (e) \(-\tan t\), (f) \(2\sin 2t\), (g) \(-\csc t \cot t\), (h) \(2\sec^2 t \tan t\), (i) \(-e^{\cos t} \sin t\), (j) \(\tan t\), (k) \(-e^{-t} \sin t + e^{-t} \cos t\), (l) \(2\tan t \sec^2 t\), (m) \(\frac{e^{2t}}{200}\), (n) \(-20\tan t\), (o) \(\frac{1}{15t}\).

Step 1: Differentiate a) using the Product Rule

For the function \( y = 2 \sin t \cos t \), use the identity \( \sin 2t = 2 \sin t \cos t \) to rewrite the function. Therefore, \( y = \sin 2t \). The derivative of \( \sin 2t \) is \( 2 \cos 2t \) using the chain rule.

Step 2: Differentiate b) using Trigonometric Identity

The function \( y = \sin^2 t + \cos^2 t \) is a well-known trigonometric identity equal to 1. The derivative of a constant is zero.

Step 3: Differentiate c) by Applying the Chain Rule

For \( y = \sec t = \frac{1}{\cos t} \), use the derivative of secant: \( \frac{d}{dt} \sec t = \sec t \tan t \).

Step 4: Differentiate d) using the Quotient Rule

Given \( y = \cot t \), which is \( \frac{\cos t}{\sin t} \), apply the quotient rule: \( \frac{du}{dt} = \frac{-\sin^2 t - \cos^2 t}{\sin^2 t} = -\csc^2 t \).

Step 5: Differentiate e) using Chain Rule

The derivative of \( y = \ln \cos t \) is \( \frac{-\sin t}{\cos t} = -\tan t \) using the chain rule.

Step 6: Differentiate f) using Trigonometric Identities

\( y = \sin^2 t - \cos^2 t \) simplifies using the identity for \( \cos 2t = 2 \cos^2 t - 1 \). Therefore, it becomes \( -\cos 2t \). The derivative is \( 2 \sin 2t \).

Step 7: Differentiate g) using Chain Rule

For \( y = \csc t = \frac{1}{\sin t} \), use \( \frac{d}{dt} \csc t = -\csc t \cot t \).

Step 8: Differentiate h) using Chain Rule

The function \( y = \sec^2 t \) is derived using the chain rule: \( 2 \sec t \sec t \tan t = 2 \sec^2 t \tan t \).

Step 9: Differentiate i) using Chain Rule

For \( y = e^{\cos t} \), the derivative is \( -e^{\cos t} \sin t \) using the chain rule.

Step 10: Differentiate j) using Chain Rule

Given \( y = \ln (\sec t) \), use the identity \( \sec t = \frac{1}{\cos t} \). The derivative is \( \tan t \).

Step 11: Differentiate k) using Product Rule

For \( y = e^{-t} \sin t \), differentiate:\( y' = -e^{-t} \sin t + e^{-t} \cos t \) using the product rule.

Step 12: Differentiate l) using Chain Rule

Given \( y = \tan^2 t \), let \( u = \tan t \). So, \( y = u^2 \). The derivative is \( 2 \tan t \sec^2 t \).

Step 13: Differentiate m) Directly

The function \( y = \frac{e^{2t}}{400} \) has a derivative \( y' = \frac{2e^{2t}}{400} = \frac{e^{2t}}{200} \) since constants factor out.

Step 14: Differentiate n) using Chain Rule

The function \( y=\ln(\cos^{20}t) \) requires applying the chain rule: \( 20 (\ln(\cos t))' = 20 \left(\frac{-\sin t}{\cos t} \right) = -20\tan t \).

Step 15: Differentiate o) using Chain Rule

For \( y = \frac{\ln t^2}{30} \), simplify to \( \frac{2 \ln t}{30} \), so \( y' = \frac{2}{30t} = \frac{1}{15t} \).

Key Concepts

Trigonometric Derivatives

Trigonometric functions, like sine, cosine, and tangent, have specific derivatives that are invaluable in calculus. Differentiating these functions requires you to memorize and apply certain rules.
- The derivative of \( \sin t \) is \( \cos t \).
- The derivative of \( \cos t \) is \( -\sin t \).
- For \( \tan t \), the derivative is \( \sec^2 t \).

These rules are foundational. Once they are known, you can use them in combination with other rules, like the chain rule, to handle more complex functions. For instance, differentiating functions like \( \sec t = \frac{1}{\cos t} \) or \( \csc t = \frac{1}{\sin t} \) elegantly demonstrates how these basic derivatives are used.

Chain Rule

The chain rule is a crucial tool in calculus, especially when you are dealing with composite functions — these are functions nested inside another function. The chain rule helps you find the derivative of these complex structures by breaking them down.
For example, for the function \( y = e^{\cos t} \), think of \( y \) in terms of two layers: the outer function \( e^u \, \) where \( u = \cos t \. \) You first find the derivative of the outer function and multiply it by the derivative of the inner function: \( -e^{\cos t} \sin t \. \)

• Differentiate the outer function.
• Multiply it by the derivative of the inner function.

This method lets you tackle derivatives efficiently without losing sight of the underlying structure of the function.

Product Rule

Whenever you face a problem related to the differentiation of two multiplicative functions, the product rule comes into play. This rule states that if you have a function in the form \( y = u(t) \, v(t) \), then the derivative is given by \( y' = u'v + uv' \).
Take, for example, the problem where \( y = e^{-t} \sin t \. \) Applying the product rule, the derivative is:
\[ y' = (-e^{-t})(\sin t) + (e^{-t})(\cos t). \]
Here’s a step-by-step fashion to apply the product rule:

• Differentiate the first function while keeping the second one constant.
• Add the product of the first function, unchanged, and the derivative of the second function.

By systematically applying this rule, you handle complex multiplicative derivatives.

Quotient Rule

The quotient rule is essential for situations where you have one function divided by another. Its formula helps tackle these fractions directly. To derive a function \( y = \frac{u}{v} \, \) the quotient rule states that the derivative \( y' = \frac{u'v - uv'}{v^2} \. \)
Consider the function \( y = \cot t = \frac{\cos t}{\sin t} \. \) Using the quotient rule:
\[ y' = \frac{(-\sin t)(\sin t) - (\cos t)(\cos t)}{\sin^2 t} = -\csc^2 t. \]
The quotient rule can be demystified by these simple steps:

• Take the derivative of the numerator, multiplying it by the denominator as is.
• Subtract the product of the numerator and the derivative of the denominator.
• Divide the entire expression by the square of the denominator.

Understanding these steps allows you to approach division-related differentiation with confidence.