Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find \(y^{\prime}\) for a. \(y=2 \cos t\) b. \(y=\cos 2 t\) c. \(y=\sin t^{2}\) d. \(y=\sin (t+\pi)\) e. \(y=\cos (\pi t-\pi / 2)\) f. \(y=(\sin t) \times(\cos t)\) g. \(\quad y=\sin ^{2}\left(t^{4}\right)\) h. \(y=\cos (\ln (t+1))\) i. \(y=\sin (\cos t)\) j. \(y=\tan \left(\frac{\pi}{2} t\right)\) k. \(y=\tan ^{2}\left(t^{2}\right) \quad\) l. \(y=\tan (\cos t)\) \(\mathrm{m} \cdot y=-\ln (\cos t)\) n. \(y=e^{\sin t}\) o. \(y=\ln (\sin t)\)

Step-by-Step Solution

Answer

a. \, -2\sin t \, | \, b. \, -2\sin(2t) \, | \, c. \, 2t\cos(t^2) \, | \, d. \, -\cos t \, | \, e. \, -\pi\sin(\pi t - \pi/2) \, | \, f. \, \cos^2 t - \sin^2 t \, | \, g. \, 8t^3\sin(t^4)\cos(t^4) \, | \, h. \, -\sin(\ln(t+1))/(t+1) \, | \, i. \, -\sin t\cos(\cos t) \, | \, j. \, (\pi/2)\sec^2(\pi t/2) \, | \, k. \, 4t\tan(t^2)\sec^2(t^2) \, | \, l. \, -\sin t\sec^2(\cos t) \, | \, m. \, \tan t \, | \, n. \, e^{\sin t}\cos t \, | \, o. \, \cot t

Step 1: Differentiate Part (a)

Given: \( y = 2 \cos t \). The derivative of \( \cos t \) is \( -\sin t \). Therefore, \( y' = 2 \cdot (-\sin t) = -2 \sin t \).

Step 2: Differentiate Part (b)

Given: \( y = \cos 2t \). Using the chain rule, \( y' = -\sin(2t) \cdot (2) = -2\sin(2t) \).

Step 3: Differentiate Part (c)

Given: \( y = \sin t^2 \). Using the chain rule, \( y' = \cos(t^2) \cdot (2t) = 2t\cos(t^2) \).

Step 4: Differentiate Part (d)

Given: \( y = \sin(t+\pi) \). Since \( \sin(t+\pi) = -\sin t \), then \( y' = -\cos t \).

Step 5: Differentiate Part (e)

Given: \( y = \cos(\pi t - \pi / 2) \). Using the chain rule, \( y' = -\sin(\pi t - \pi / 2) \cdot (\pi) = -\pi \sin(\pi t - \pi / 2) \).

Step 6: Differentiate Part (f)

Given: \( y = (\sin t) \times (\cos t) \). Apply the product rule: \( y' = \sin t \cdot (-\sin t) + \cos t \cdot \cos t = \cos^2 t - \sin^2 t \).

Step 7: Differentiate Part (g)

Given: \( y = \sin^2(t^4) \). Apply the chain rule: \( y' = 2\sin(t^4)\cos(t^4)\cdot (4t^3) = 8t^3 \sin(t^4)\cos(t^4) \).

Step 8: Differentiate Part (h)

Given: \( y = \cos(\ln(t+1)) \). Using the chain rule: \( y' = -\sin(\ln(t+1)) \cdot \frac{1}{t+1} \).

Step 9: Differentiate Part (i)

Given: \( y = \sin(\cos t) \). Using the chain rule: \( y' = \cos(\cos t) \cdot (-\sin t) = -\sin t \cdot \cos(\cos t) \).

Step 10: Differentiate Part (j)

Given: \( y = \tan(\frac{\pi}{2} t) \). Using the chain rule: \( y' = \sec^2(\frac{\pi}{2} t) \cdot (\frac{\pi}{2}) \).

Step 11: Differentiate Part (k)

Given: \( y = \tan^2(t^2) \). Apply the chain rule: \( y' = 2\tan(t^2) \cdot \sec^2(t^2) \cdot (2t) = 4t\tan(t^2)\sec^2(t^2) \).

Step 12: Differentiate Part (l)

Given: \( y = \tan(\cos t) \). Using the chain rule: \( y' = \sec^2(\cos t) \cdot (-\sin t) = -\sin t \sec^2(\cos t) \).

Step 13: Differentiate Part (m)

Given: \( y = -\ln(\cos t) \). Using the chain rule: \( y' = -\frac{1}{\cos t} \cdot (-\sin t) = \tan t \).

Step 14: Differentiate Part (n)

Given: \( y = e^{\sin t} \). Using the chain rule: \( y' = e^{\sin t} \cdot \cos t \).

Step 15: Differentiate Part (o)

Given: \( y = \ln(\sin t) \). Using the chain rule: \( y' = \frac{1}{\sin t} \cdot \cos t = \cot t \).

Key Concepts

Chain Rule

The chain rule is a powerful technique in calculus used to differentiate compositions of functions. Let's break it down.

• When you have a function nested within another function, the chain rule helps to find the derivative.
• Imagine a scenario where a variable, say, \( u \), is expressed in terms of another variable \( t \), such that \( y = f(u) \) and \( u = g(t) \). According to the chain rule, the derivative of \( y \) with respect to \( t \) is the derivative of \( y \) with respect to \( u \) multiplied by the derivative of \( u \) with respect to \( t \).

Expressing this using math notation:
\[ y' = \frac{dy}{du} \times \frac{du}{dt} \]
This rule simplifies the process of differentiation, especially for complex composites like \( y = \sin(t^2) \). Here, we obtain \( y' = 2t \cos(t^2) \) by setting \( u = t^2 \), differentiating \( \sin(u) \) and \( u(t) \) separately, and multiplying the results.

Product Rule

The product rule is essential when you need to differentiate the product of two functions. This rule is particularly useful when two functions are multiplied together, and you need their combined effect in a derivative.

• For instance, consider two functions \( u(t) \) and \( v(t) \).
• The product rule states that the derivative of their product, \( y = u(t) \times v(t) \), is given by \( y' = u' \times v + u \times v' \).

Here's how it applies to \( y = (\sin t) \times (\cos t) \):

• First, find the derivatives of the individual functions: \( u' = \cos t \) and \( v' = -\sin t \).
• Then apply the product rule: \( y' = (\sin t)(-\sin t) + (\cos t)(\cos t) \), which simplifies to \( y' = \cos^2 t - \sin^2 t \).

This formula incorporates the essence of both the chain and product rules where needed, making it versatile for handling complex differentiations.

Trigonometric Functions

Trigonometric functions are foundational in calculus, often used as an inherent element in many problems.

• Basic functions like \( \sin t \), \( \cos t \), and \( \tan t \) have simple derivatives: \( \sin t \) differentiates to \( \cos t \); \( \cos t \) differentiates to \( -\sin t \); and \( \tan t \) differentiates to \( \sec^2 t \).
• These derivatives are crucial for solving more intricate mathematical derivatives.

A typical example would be differentiating \( y = \cos(\ln(t+1)) \):

• Use the chain rule where the outer function is a cosine and the inner function is a natural logarithm.
• The derivative is \( y' = -\sin(\ln(t+1)) \cdot \frac{1}{t+1} \).

Understanding these basic trigonometric derivatives allows for the tackling of more complex functions, such as those involving compositions of trigonometric expressions and other non-linear functions.