Problem 17 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=\cos \left(e^{-\theta^{2}}\right)\end{equation}

Step-by-Step Solution

Answer

The derivative of the function with respect to \( \theta \) is \( \frac{dy}{d\theta} = 2\theta \sin(e^{-\theta^2}) e^{-\theta^2} \).

Step 1: Identify the function

The given function is \( y = \cos(e^{-\theta^2}) \). We need to find the derivative of this function with respect to \( \theta \).

Step 2: Apply the chain rule

To differentiate \( y = \cos(e^{-\theta^2}) \) with respect to \( \theta \), we use the chain rule. The chain rule states that if \( y = f(g(\theta)) \), then \( \frac{dy}{d\theta} = f'(g(\theta)) \cdot g'(\theta) \).

Step 3: Differentiate the outer function

Let \( u = e^{-\theta^2} \), so our function is transformed into \( y = \cos(u) \). The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \).

Step 4: Differentiate the inner function

Now differentiate \( u = e^{-\theta^2} \) with respect to \( \theta \). The derivative of \( e^{v} \) is \( e^{v} \) times the derivative of \( v \) with respect to \( \theta \), where \( v = -\theta^2 \). Thus, \( \frac{d}{d\theta}(e^{-\theta^2}) = e^{-\theta^2} \cdot (-2\theta) \).

Step 5: Combine results using chain rule

Using the chain rule, the derivative of \( y = \cos(e^{-\theta^2}) \) with respect to \( \theta \) is \( \frac{dy}{d\theta} = -\sin(e^{-\theta^2}) \cdot (e^{-\theta^2} \cdot (-2\theta)) \).

Step 6: Simplify the expression

Simplifying the expression gives \( \frac{dy}{d\theta} = 2\theta \sin(e^{-\theta^2}) e^{-\theta^2} \).

Key Concepts

Understanding the Chain Rule

The chain rule is a crucial concept in calculus used when differentiating composite functions. Imagine you have a function nested inside another function. The chain rule helps us find the derivative of such functions efficiently. In more technical terms, if you have a function composed of two functions, say \( y = f(g(x)) \), you need to differentiate both.

• Differentiating \( f \) gives \( f'(g(x)) \).
• Differentiating \( g \) gives \( g'(x) \).
• The chain rule tells us that the derivative of the composite function is \( f'(g(x)) \times g'(x) \).

Next time you see a function within another, remember the chain rule is your best friend!

Trigonometric Differentiation Essentials

Trigonometric differentiation is all about finding derivatives of trig functions like sine, cosine, tangent, etc. Each of these functions has its own unique derivative:

• The derivative of \( \cos(u) \) is \( -\sin(u) \).
• The derivative of \( \sin(u) \) is \( \cos(u) \).
• The derivative of \( \tan(u) \) is \( \sec^2(u) \).

These derivatives are very useful in solving calculus problems, especially those involving composite functions like in our original exercise. Remember, if you transform your function into a trigonometric one, always apply these basic rules while differentiating.

The Power of Exponential Functions

Exponential functions are another common topic in calculus. Functions such as \( e^{x} \) have their own special properties. The derivative of an exponential function of the form \( e^{v} \) is, quite interestingly, \( e^{v} \times \frac{dv}{dx} \).When differentiating \( e^{v} \), where \( v \) is a function of another variable like \( x \) or \( \theta \), we multiply the whole expression by the derivative of \( v \). This is particularly useful when applying the chain rule in more complex functions involving exponentials. These functions grow extremely fast, and their derivatives help in capturing this growth.

Exploring Multivariable Calculus

Multivariable calculus extends single-variable calculus principles to functions with more than one independent variable. An example is \( z = f(x, y) \), where you have partial derivatives: derivatives taken with respect to one variable while holding others constant.

• Partial differentiation lets us see how changes in each variable affect the function while other variables remain fixed.
• For example, \( \frac{\partial z}{\partial x} \) explores how \( z \) changes as \( x \) changes, keeping \( y \) constant.
• This concept is vital for fields like physics or engineering, where multiple variables often influence an outcome.

The transition from single-variate to multivariable calculus involves understanding these partial changes, providing a more comprehensive picture of how functions behave in various conditions.