Proof Of The Derivative Of Sinx

Diving into the derivative of sin(x) is like embarking on a journey through the heart of calculus, where we uncover the elegant relationship between a function and its rate of change. Grasping this derivative isn't just about memorizing a formula; it's about understanding the fundamental principles that underpin calculus itself.

The Essence of Derivatives

At its core, a derivative measures how a function changes as its input changes. Imagine driving a car: your speed (the rate at which your position changes) is the derivative of your position function. Similarly, the derivative of sin(x) tells us how the sine function changes as x changes.

The Limit Definition of a Derivative

The journey to proving the derivative of sin(x) begins with the limit definition of a derivative:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

This formula calculates the instantaneous rate of change of a function f(x) at a specific point x. We’ll use this to find the derivative of sin(x).

Setting the Stage: sin(x) and its Derivative

Our mission is to find the derivative of f(x) = sin(x). Applying the limit definition, we get:

f'(x) = lim(h->0) [sin(x + h) - sin(x)] / h

This expression might look daunting, but we'll break it down step by step.

The Sine Addition Formula: Our Key Tool

To simplify sin(x + h), we use the sine addition formula:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

Applying this to our expression, we have:

sin(x + h) = sin(x)cos(h) + cos(x)sin(h)

Substituting and Simplifying

Now, substitute this back into our limit definition:

f'(x) = lim(h->0) [sin(x)cos(h) + cos(x)sin(h) - sin(x)] / h

Rearrange the terms to group the sin(x) terms together:

f'(x) = lim(h->0) [sin(x)cos(h) - sin(x) + cos(x)sin(h)] / h

Factor out sin(x) from the first two terms:

f'(x) = lim(h->0) [sin(x)(cos(h) - 1) + cos(x)sin(h)] / h

Now, split the limit into two separate limits:

f'(x) = lim(h->0) [sin(x)(cos(h) - 1) / h] + lim(h->0) [cos(x)sin(h) / h]

Leveraging Limit Properties

Since sin(x) and cos(x) do not depend on h, we can pull them out of the limits:

f'(x) = sin(x) * lim(h->0) [(cos(h) - 1) / h] + cos(x) * lim(h->0) [sin(h) / h]

Now, we need to evaluate these two crucial limits:

lim(h->0) [(cos(h) - 1) / h] and lim(h->0) [sin(h) / h]

Evaluating lim(h->0) [sin(h) / h]

This limit is a classic in calculus. We'll demonstrate it using a geometric argument.

Geometric Proof

1. Consider a Unit Circle: Imagine a circle with radius 1 centered at the origin of a coordinate plane.

2. Define Angle h: Let h be a small positive angle in radians, measured counterclockwise from the positive x-axis.

3. Areas of Sectors and Triangles:

   • Consider the triangle formed by the origin, the point (1,0) on the x-axis, and the point on the unit circle corresponding to the angle h. The area of this triangle is (1/2) * base * height = (1/2) * 1 * sin(h) = (1/2)sin(h).
   • Consider the sector of the circle enclosed by the angle h. The area of this sector is (1/2) * r^2 * h = (1/2) * 1^2 * h = (1/2)h.
   • Consider the triangle formed by the origin, the point (1,0), and the point on the vertical line x=1 that intersects the extension of the line from the origin through the point on the unit circle corresponding to angle h. The height of this triangle is tan(h), so its area is (1/2) * base * height = (1/2) * 1 * tan(h) = (1/2)tan(h).

4. Area Inequality: From the geometry, we have the following inequality:

   Area of triangle < Area of sector < Area of larger triangle (1/2)sin(h) < (1/2)h < (1/2)tan(h)

5. Simplify the Inequality: Multiply all parts of the inequality by 2:

   sin(h) < h < tan(h)

6. Divide by sin(h): Since h is a small positive angle, sin(h) is positive. We can divide all parts of the inequality by sin(h) without changing the direction of the inequalities:

   1 < h/sin(h) < tan(h)/sin(h) 1 < h/sin(h) < (sin(h)/cos(h))/sin(h) 1 < h/sin(h) < 1/cos(h)

7. Take the Reciprocal: Taking the reciprocal of each part reverses the direction of the inequalities:

   1 > sin(h)/h > cos(h) cos(h) < sin(h)/h < 1

8. Apply the Squeeze Theorem: As h approaches 0, cos(h) approaches 1. Therefore, by the Squeeze Theorem:

   lim(h->0) cos(h) = 1 lim(h->0) 1 = 1 Thus, lim(h->0) sin(h)/h = 1

L'Hôpital's Rule Proof

1. Indeterminate Form: As h approaches 0, sin(h) approaches 0, and h approaches 0. Thus, the limit is of the indeterminate form 0/0.

2. Apply L'Hôpital's Rule: L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches c is of the form 0/0 or ∞/∞, then:

   lim(x->c) [f(x)/g(x)] = lim(x->c) [f'(x)/g'(x)]

3. Differentiate: Differentiate the numerator and the denominator:

   f(h) = sin(h) => f'(h) = cos(h) g(h) = h => g'(h) = 1

4. Evaluate the Limit:

   lim(h->0) [sin(h)/h] = lim(h->0) [cos(h)/1] As h approaches 0, cos(h) approaches 1.

   lim(h->0) [cos(h)/1] = cos(0)/1 = 1/1 = 1 Therefore, lim(h->0) [sin(h)/h] = 1

Evaluating lim(h->0) [(cos(h) - 1) / h]

This limit requires a bit more manipulation. We'll use a clever algebraic trick.

Algebraic Manipulation and the Squeeze Theorem

1. Multiply by the Conjugate: Multiply the numerator and denominator by the conjugate of the numerator, which is (cos(h) + 1):

   lim(h->0) [(cos(h) - 1) / h] = lim(h->0) [(cos(h) - 1)(cos(h) + 1) / (h(cos(h) + 1))]

2. Simplify:

   • Expand the numerator using the difference of squares: (cos(h) - 1)(cos(h) + 1) = cos^2(h) - 1.
   • Recall the Pythagorean identity: sin^2(h) + cos^2(h) = 1. Thus, cos^2(h) - 1 = -sin^2(h).

   lim(h->0) [-sin^2(h) / (h(cos(h) + 1))]

3. Rewrite the Limit:

   • Separate the limit into two parts:

   lim(h->0) [-sin(h) / h] * lim(h->0) [sin(h) / (cos(h) + 1)]

4. Evaluate the Limits:

   • We know lim(h->0) [sin(h) / h] = 1.
   • As h approaches 0, sin(h) approaches 0, and cos(h) approaches 1:

   lim(h->0) [sin(h) / (cos(h) + 1)] = 0 / (1 + 1) = 0 / 2 = 0

5. Combine the Results:

   lim(h->0) [(cos(h) - 1) / h] = -1 * 0 = 0

L'Hôpital's Rule Proof

1. Indeterminate Form: As h approaches 0, cos(h) approaches 1, so cos(h) - 1 approaches 0. Also, h approaches 0. Thus, the limit is of the indeterminate form 0/0.

2. Apply L'Hôpital's Rule:

   • L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches c is of the form 0/0 or ∞/∞, then:

     lim(x->c) [f(x)/g(x)] = lim(x->c) [f'(x)/g'(x)]

3. Differentiate:

   • Differentiate the numerator and the denominator:

     f(h) = cos(h) - 1 => f'(h) = -sin(h) g(h) = h => g'(h) = 1

4. Evaluate the Limit:

   lim(h->0) [(cos(h) - 1) / h] = lim(h->0) [-sin(h) / 1] As h approaches 0, sin(h) approaches 0.

   lim(h->0) [-sin(h) / 1] = -sin(0) / 1 = 0 / 1 = 0 Therefore, lim(h->0) [(cos(h) - 1) / h] = 0.

Putting It All Together

Now that we've evaluated the limits, we can substitute them back into our expression for f'(x):

f'(x) = sin(x) * lim(h->0) [(cos(h) - 1) / h] + cos(x) * lim(h->0) [sin(h) / h] f'(x) = sin(x) * 0 + cos(x) * 1 f'(x) = 0 + cos(x) f'(x) = cos(x)

Thus, the derivative of sin(x) is cos(x).

Alternative Proof using Euler's Formula

Another elegant proof involves using Euler's formula, which connects complex exponentials to trigonometric functions:

e^(ix) = cos(x) + i*sin(x)

where i is the imaginary unit (i^2 = -1).

Expressing sin(x) using Euler's Formula

From Euler's formula, we can express sin(x) as:

sin(x) = (e^(ix) - e^(-ix)) / (2i)

Differentiating sin(x)

Now, let's differentiate sin(x) with respect to x:

d/dx [sin(x)] = d/dx [(e^(ix) - e^(-ix)) / (2i)]

Using the linearity of differentiation, we get:

d/dx [sin(x)] = (1 / (2i)) * [d/dx (e^(ix)) - d/dx (e^(-ix))]

Now, we need to differentiate e^(ix) and e^(-ix).

Differentiating e^(ix) and e^(-ix)

Using the chain rule, we have:

d/dx (e^(ix)) = i * e^(ix) d/dx (e^(-ix)) = -i * e^(-ix)

Substituting Back

Substitute these derivatives back into our expression:

d/dx [sin(x)] = (1 / (2i)) * [i * e^(ix) - (-i) * e^(-ix)] d/dx [sin(x)] = (1 / (2i)) * [i * e^(ix) + i * e^(-ix)] d/dx [sin(x)] = (i / (2i)) * [e^(ix) + e^(-ix)] d/dx [sin(x)] = (1 / 2) * [e^(ix) + e^(-ix)]

Recognizing cos(x)

Recall Euler's formula again:

cos(x) = (e^(ix) + e^(-ix)) / 2

Therefore, our expression simplifies to:

d/dx [sin(x)] = cos(x)

This completes the proof using Euler's formula.

Visualizing the Derivative

Understanding the derivative of sin(x) becomes more intuitive with a visual representation. Imagine the graph of sin(x). At points where the graph has a positive slope, cos(x) is positive; where sin(x) has a negative slope, cos(x) is negative; and where sin(x) has a zero slope (peaks and troughs), cos(x) is zero. This visual correlation reinforces the relationship between sin(x) and its derivative, cos(x).

Practical Applications

The derivative of sin(x) isn't just a theoretical concept. It has numerous applications in physics, engineering, and mathematics. For instance, in physics, it's used to describe simple harmonic motion, such as the motion of a pendulum or a mass on a spring. In signal processing, it's used to analyze and manipulate sinusoidal signals.

Common Questions and Misconceptions

• Why is the limit definition necessary? The limit definition provides a rigorous foundation for calculating derivatives. It allows us to find the instantaneous rate of change, which is crucial for many applications.
• Can I just memorize the derivative? While memorization can be helpful, understanding the derivation provides a deeper understanding of the concept and allows you to apply it in various contexts.
• What if x is in degrees? If x is in degrees, you need to convert it to radians before applying the derivative. The derivative of sin(x) is cos(x) only when x is in radians.
• Is there a simpler way to think about this? Visualizing the graphs of sin(x) and cos(x) and understanding their relationship can make the concept more intuitive.

Conclusion

Proving the derivative of sin(x) is a rewarding exercise that reinforces the core principles of calculus. Whether using the limit definition, geometric arguments, or Euler's formula, each approach provides a unique perspective on this fundamental concept. By understanding this derivative, you unlock a deeper understanding of calculus and its applications in various fields. The journey from the limit definition to the final result, d/dx [sin(x)] = cos(x), is a testament to the elegance and power of mathematical reasoning.