Problem 12 - Calculus

Question

Compute the slope at \(x=0\) directly from limits: (a) \(y=\tan x\) (b) \(y=\sin (-x)\)

Step-by-Step Solution

Answer

For (a), the slope at \(x=0\) is 1; for (b), it is -1.

Step 1: Understand the Problem

We need to find the derivative of the given functions at the point where \(x=0\). This is done by calculating the limits of the difference quotient \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).

Step 2: Differentiate \(y=\tan x\)

For \(y = \tan x\), the derivative is defined as \(f'(x) = \sec^2 x\). We need to find \(f'(0)\).Let's calculate:\[f'(0) = \sec^2(0) = (\cos(0))^{-2} = 1^2 = 1\]

Step 3: Confirm \(y=\tan x\) by Limits

Using the limit definition, substitute \(f(x) = \tan x\): \[ \lim_{h \to 0} \frac{\tan(h) - \tan(0)}{h} = \lim_{h \to 0} \frac{\tan(h)}{h}\]Knowing that \( \tan(h) \approx h \) when \( h \) is close to 0, the limit becomes:\[ \lim_{h \to 0} \frac{h}{h} = 1 \] confirming the result from calculus.

Step 4: Differentiate \(y=\sin(-x)\)

For \(y = \sin(-x)\), apply the chain rule: the derivative is \(-\cos(x)\).Calculate the slope at \(x=0\):\[y'(0) = -\cos(0) = -1\].

Step 5: Confirm \(y=\sin(-x)\) by Limits

Using limits, we substitute \(f(x) = \sin(-x)\): \[ \lim_{h \to 0} \frac{\sin(-h) - \sin(0)}{h} = \lim_{h \to 0} \frac{-\sin(h)}{h}\]Since \(\lim_{h \to 0} \frac{\sin(h)}{h} = 1\), it follows:\[\lim_{h \to 0} \frac{-\sin(h)}{h} = -1\], confirming the derivative.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function changes at any given point. This rate is known as the derivative. The derivative tells us the slope of the tangent line to the curve of a function at a specific point.When you differentiate a function, you're essentially looking for how one variable changes in relation to another. In simpler terms, differentiation helps in understanding how things change.The process often begins with the difference quotient:

• Consider the function as it changes by a small amount, usually denoted as \( h \).
• This gives us the expression \( \frac{f(x+h) - f(x)}{h} \).
• By finding the limit as \( h \to 0 \), we get the derivative \( f'(x) \). This is the exact change rate we are looking for.

Differentiation isn't just about formulas. It requires understanding when and how different rules apply, such as the product rule, quotient rule, and chain rule.

Limits

Limits are the core of calculus. They help us understand the behavior of functions as they approach a certain point. Whether we're dealing with functions that explode to infinity or settle into a steady value, limits give us precise definitions.When calculating derivatives using limits, remember:

• The limit process considers what happens to a function as it gets infinitely close to a certain point, say \( x \) becoming \( x + h \).
• For derivatives, we look at how the function behaves as \( h \to 0 \).

In the given exercise, to find the slope of \( y = \tan x \) and \( y = \sin(-x) \) at \( x = 0 \), limits were used to confirm the derivatives. This method reassures that differentiation rules provide correct results, mirroring the actual behavior of functions at specific points.

Trigonometric Functions

Trigonometric functions, like \( \tan x \) and \( \sin x \), are periodic and arise often in calculus. They have unique properties that affect how they are differentiated.

• For \( y = \tan x \), the derivative is \( \sec^2 x \).
• At \( x = 0 \), this simplifies to \( 1 \) because \( \cos(0) = 1 \).
• For \( y = \sin(-x) \), differentiation using the chain rule gives \( -\cos x \).
• At \( x = 0 \), we find the slope is \( -1 \) as \( \cos(0) = 1 \).

Understanding these derivatives is crucial because trigonometric functions frequently serve as models for periodic phenomena such as waves, oscillations, and other cycles. They showcase how changes occur not just in linear patterns, but in repeating cycles that are predictable and calculable.