Differentiation of cos(x)

The differentiation of (\cos(x)) with respect to (x) is a fundamental concept in calculus. The derivative of (\cos(x)) is given by:

\[
\frac{d}{dx}(\cos(x)) = -\sin(x).
\]

This result is derived from the limit definition of the derivative and is a key part of the differentiation rules for trigonometric functions. The negative sine function, $-\sin(x)$, represents the rate of change of $\cos(x)$ with respect to (x).

Proof of Formula

We can proof using the below two formulas

$\frac {d}{dx} f(x) =\displaystyle \lim_{h \to 0} \frac {f (x+h) – f(x)}{h}$

$\displaystyle \lim_{x \to 0} \frac {\sin(x)}{x}=1$

Now

$\frac {d}{dx} (\cos x) = \displaystyle \lim_{h \to 0} \frac {\cos (x+h) – \cos x}{h}= \displaystyle \lim_{h \to 0} -\frac {2\sin (\frac {2x+h}{2}) sin \frac {h}{2}}{h}$

$=-\displaystyle \lim_{h \to 0} \sin (x + \frac {h}{2}) \displaystyle \lim_{h \to 0} \frac {sin \frac {h}{2}}{\frac {h}{2}} =- \sin x$

as $\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$

Domain of the Derivative of cosx

\[
\frac{d}{dx}(\cos(x)) =- \sin(x).
\]

We can clearly see that Domain of the Derivative of cos x is R

Geometrical Meaning of the Derivative

The geometric meaning of the derivative of $\cos(x)$, which is $-\sin(x)$, can be understood by examining the graph of $\cos(x)$ and considering what the derivative represents.

1. Slope of the Tangent Line: The derivative of a function at a point gives the slope of the tangent line to the graph of the function at that point. For cos(x), the derivative at any point (x) is -sin(x). This means that the slope of the tangent line to the curve y = cos(x) at any point (x) is equal to -sin(x).
2. At the Peaks and Troughs: At the points where $\cos(x)$ reaches its maximum or minimum (i.e., where the curve is flattest), the derivative (slope) is $0$ because $\sin(x)$ is $0$ at $x = n\pi$ where $n$ is an integer. This aligns with the geometric intuition that at the top and bottom of a wave, the slope of the tangent line (indicating the rate of change) is horizontal.
3. Zero Crossings of cos(x): At the points where $\cos(x)$ crosses the x-axis (i.e., $\cos(x) = 0$), the derivative $-\sin(x)$ takes its maximum absolute value ($-1$ or $1$), indicating that the slope of the tangent to the curve is steepest at these points. This is because $\sin(x) = \pm 1$ at $x = \frac{\pi}{2} + n\pi$, corresponding to the steepest ascent or descent of the $\cos(x)$ curve.
4. Direction of the Slope: The negative sign in $-\sin(x)$ indicates that as $x$ increases, the function $\cos(x)$ generally decreases when $\sin(x)$ is positive, and increases when $\sin(x)$ is negative. This negative relationship reflects the downward and upward slopes of the curve respectively.

Solved Questions

Question 1:
Differentiate $y = \cos(3x)$.
Solution:
To differentiate $y = \cos(3x)$, we use the chain rule. Let (u = 3x), then $y = \cos(u)$. Differentiating (y) with respect to (x):
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin(u) \cdot 3 = -3\sin(3x).
\]

Question 2:
Find the derivative of $y = x \cos(x)$.
Solution
Here, we use the product rule. If (y = uv) where (u = x) and $v = \cos(x)$, then (y’ = u’v + uv’). Therefore:

\[
\frac{dy}{dx} = 1 \cdot \cos(x) – x \cdot \sin(x) = \cos(x) – x\sin(x).
\]

Question 3:
What is the second derivative of $y = \cos(x)$?
Solution:
First, we find the first derivative, which is $y’ = -\sin(x)$. Then, differentiating again:

\[
y” = \frac{d}{dx}(\cos(x)) = -\cos(x).
\]

Question 4
Find the equation of the tangent line to the curve $y = \cos(x)$ at $x = \frac{\pi}{4}$.

Solution
First, find the derivative of $y = \cos(x)$, which is $y’ = -\sin(x)$.
Then, evaluate this derivative at $x = \frac{\pi}{4}$ to find the slope of the tangent line:
$$y'(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$

The slope of the tangent line at $x = \frac{\pi}{4}$ is $-\frac{\sqrt{2}}{2}$. The y-coordinate at this point is $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. Using the point-slope form of the equation of a line:
$$y – \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}(x – \frac{\pi}{4})$$

Question 5
Determine the $x$ values where the function $f(x) = \cos(x)$ has maximum and minimum values on the interval $[0, 2\pi]$.

Solution
The derivative of $f(x) = \cos(x)$ is $f'(x) = -\sin(x)$. Setting the derivative equal to 0 to find critical points:
$$-\sin(x) = 0$$

Solving for $x$, we get:
$$x = n\pi, \text{ where } n \text{ is an integer}$$

Within the interval $[0, 2\pi]$, the relevant values of $x$ are $0$, $\pi$, and $2\pi$. At $x = 0$ and $x = 2\pi$, $\cos(x) = 1$, indicating maximum values. At $x = \pi$, $\cos(x) = -1$, indicating a minimum value.

Question 6
Prove that the derivative of $\sin(x)$ is $\cos(x)$ using the derivative of $\cos(x)$.

Solution

Knowing that $\cos(x) = \sin(x + \frac{\pi}{2})$, we can differentiate both sides with respect to $x$:
$$f(x) = \sin(x + \frac{\pi}{2})$$

The derivative of the right side using the chain rule is:
$$f'(x) = \cos(x + \frac{\pi}{2}) \cdot \frac{d}{dx}(x + \frac{\pi}{2}) = \cos(x + \frac{\pi}{2})$$

Since $\cos(x + \frac{\pi}{2}) = \cos(x)$’s derivative is $-\sin(x)$, and by the periodicity and phase shift properties of sine and cosine, this demonstrates the relationship and the derivative of $\sin(x)$ indirectly.