Derivative Of Cot(x) At X=π/2: A Step-by-Step Solution

Hey there, math enthusiasts! Ever wondered how to find the derivative of cotangent and evaluate it at a specific point? Today, we're diving into a classic calculus problem: finding the value of $\frac{dy}{dx}$ for $y = \cot(x)$ when $x = \frac{\pi}{2}$. Don't worry, we'll break it down step-by-step so it's super easy to follow. Whether you're a student tackling homework or just brushing up on your calculus skills, this guide is for you. So, let's get started and unravel this mathematical puzzle together!

Understanding the Problem

Before we jump into solving, let's make sure we understand what the question is asking. We're given the function $y = \cot(x)$, and we need to find its derivative, which is $\frac{dy}{dx}$. The derivative tells us the instantaneous rate of change of the function. Once we find the derivative, we'll evaluate it at $x = \frac{\pi}{2}$ to find the slope of the cotangent function at that particular point.

What is Cotangent?

First, let's clarify what the cotangent function is. The cotangent function, often written as $\cot(x)$, is one of the fundamental trigonometric functions. It's defined as the ratio of the adjacent side to the opposite side in a right-angled triangle, or equivalently, as the reciprocal of the tangent function. Mathematically, we can express $\cot(x)$ as:

$$\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}$$

This definition is crucial because it allows us to relate cotangent to sine and cosine, which we know how to differentiate. Understanding this relationship is the first step in finding the derivative of $\cot(x)$. Remember, calculus often involves breaking down complex functions into simpler ones, and this is a perfect example of that strategy.

Why Derivatives Matter

Now, let's take a moment to appreciate why finding derivatives is so important. In calculus, the derivative of a function at a point gives us the slope of the tangent line to the function's graph at that point. This slope represents the instantaneous rate of change of the function. Imagine you're on a roller coaster; the derivative at any given moment tells you how steeply you're climbing or falling. In practical terms, derivatives are used in physics to calculate velocities and accelerations, in economics to model marginal costs and revenues, and in many other fields to understand rates of change.

In our specific problem, we're finding the rate of change of the cotangent function at $x = \frac{\pi}{2}$. This has a geometric interpretation: it's the slope of the tangent line to the graph of $y = \cot(x)$ at that point. Visualizing this can be incredibly helpful. Imagine the graph of the cotangent function, which has vertical asymptotes and decreases between them. At $x = \frac{\pi}{2}$, we're interested in how steeply the function is changing.

Finding the Derivative of cot(x)

Now that we understand the problem and the cotangent function, let's find the derivative of $y = \cot(x)$. There are a couple of ways we can approach this, and we'll explore both to give you a solid understanding.

Method 1: Using the Quotient Rule

Since $\cot(x) = \frac{\cos(x)}{\sin(x)}$, we can use the quotient rule to find its derivative. The quotient rule states that if we have a function $y = \frac{u(x)}{v(x)}$, then its derivative is given by:

$$\frac{dy}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$$

In our case, $u(x) = \cos(x)$ and $v(x) = \sin(x)$. We know the derivatives of sine and cosine:

• $u'(x) = \frac{d}{dx}(\cos(x)) = -\sin(x)$
• $v'(x) = \frac{d}{dx}(\sin(x)) = \cos(x)$

Now, we can plug these into the quotient rule formula:

$$\frac{dy}{dx} = \frac{\sin(x)(-\sin(x)) - \cos(x)(\cos(x))}{\sin^2(x)}$$

Simplify the numerator:

$$\frac{dy}{dx} = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)}$$

We can factor out a -1 from the numerator:

$$\frac{dy}{dx} = \frac{-(\sin^2(x) + \cos^2(x))}{\sin^2(x)}$$

Recall the Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$. So we have:

$$\frac{dy}{dx} = \frac{-1}{\sin^2(x)}$$

Since $\frac{1}{\sin(x)} = \csc(x)$, we can rewrite the derivative as:

$$\frac{dy}{dx} = -\csc^2(x)$$

This is a crucial result! The derivative of $\cot(x)$ is $-\csc^2(x)$. Remember this, as it's a common derivative you'll encounter in calculus.

Method 2: Using the Chain Rule

Another way to find the derivative is by using the fact that $\cot(x) = \frac{1}{\tan(x)} = [\tan(x)]^{-1}$. We can apply the chain rule here. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

In our case, let $f(u) = u^{-1}$ and $g(x) = \tan(x)$. Then, $y = f(g(x)) = [\tan(x)]^{-1}$. Now we need to find the derivatives of $f(u)$ and $g(x)$:

• $f'(u) = \frac{d}{du}(u^{-1}) = -u^{-2}$
• $g'(x) = \frac{d}{dx}(\tan(x)) = \sec^2(x)$

Applying the chain rule:

$$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = -[\tan(x)]^{-2} \cdot \sec^2(x)$$

Rewrite this as:

$$\frac{dy}{dx} = -\frac{\sec^2(x)}{\tan^2(x)}$$

Now, recall that $\sec(x) = \frac{1}{\cos(x)}$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Substitute these in:

$$\frac{dy}{dx} = -\frac{\frac{1}{\cos^2(x)}}{\frac{\sin^2(x)}{\cos^2(x)}} = -\frac{1}{\cos^2(x)} \cdot \frac{\cos^2(x)}{\sin^2(x)} = -\frac{1}{\sin^2(x)}$$

Again, since $\frac{1}{\sin(x)} = \csc(x)$, we get:

$$\frac{dy}{dx} = -\csc^2(x)$$

Both methods lead us to the same derivative, which reinforces our confidence in the result. It's always a good idea to approach a problem in multiple ways to ensure accuracy and deepen your understanding.

Evaluating the Derivative at x=π/2

Now that we've found the derivative of $\cot(x)$, which is $\frac{dy}{dx} = -\csc^2(x)$, we need to evaluate it at $x = \frac{\pi}{2}$. This means we'll substitute $\frac{\pi}{2}$ into our derivative expression.

Understanding csc(x)

Before we substitute, let's refresh our understanding of the cosecant function. The cosecant function, denoted as $\csc(x)$, is the reciprocal of the sine function:

$$\csc(x) = \frac{1}{\sin(x)}$$

To evaluate $\csc^2(\frac{\pi}{2})$, we first need to find $\sin(\frac{\pi}{2})$. Remember the unit circle? At $x = \frac{\pi}{2}$, which corresponds to 90 degrees, the sine function reaches its maximum value.

Evaluating sin(π/2)

The sine function represents the y-coordinate on the unit circle. At the angle $\frac{\pi}{2}$, the coordinates on the unit circle are (0, 1). Therefore:

$$\sin(\frac{\pi}{2}) = 1$$

This is a fundamental trigonometric value that's worth memorizing. Knowing the sine and cosine values at key angles like 0, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, and $\frac{\pi}{2}$ will make your calculus journey much smoother.

Calculating csc(π/2)

Now that we know $\sin(\frac{\pi}{2}) = 1$, we can find $\csc(\frac{\pi}{2})$:

$$\csc(\frac{\pi}{2}) = \frac{1}{\sin(\frac{\pi}{2})} = \frac{1}{1} = 1$$

So, $\csc(\frac{\pi}{2}) = 1$. This makes our final calculation straightforward.

Substituting into the Derivative

We found that $\frac{dy}{dx} = -\csc^2(x)$. Now, substitute $x = \frac{\pi}{2}$:

$$\frac{dy}{dx}|_{x=\frac{\pi}{2}} = -\csc^2(\frac{\pi}{2})$$

Since $\csc(\frac{\pi}{2}) = 1$, we have:

$$\frac{dy}{dx}|_{x=\frac{\pi}{2}} = -(1)^2 = -1$$

Therefore, the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{2}$ is -1. This means that the slope of the tangent line to the graph of $y = \cot(x)$ at $x = \frac{\pi}{2}$ is -1. The function is decreasing at this point, which aligns with the graph of the cotangent function.

Conclusion

We've successfully found the derivative of $\cot(x)$ and evaluated it at $x = \frac{\pi}{2}$. We started by understanding the problem, defining the cotangent function, and discussing the importance of derivatives. Then, we found the derivative using two different methods: the quotient rule and the chain rule, both leading us to the same result: $\frac{dy}{dx} = -\csc^2(x)$. Finally, we evaluated the derivative at $x = \frac{\pi}{2}$ and found that the value is -1.

This problem is a great example of how calculus combines trigonometric functions, differentiation rules, and algebraic manipulation. By breaking down the problem into smaller steps and understanding each concept thoroughly, we were able to arrive at the solution. Keep practicing these types of problems to build your calculus skills and confidence. Remember, the key to mastering calculus is consistent practice and a solid understanding of the fundamental concepts.

For further exploration of calculus and derivatives, you might find resources at Khan Academy's Calculus Section incredibly helpful.