Derivatives of special equations

Proof for the Derivative of $\sin(x)$ Being $\cos(x)$

Step 1: Starting with the definition of the derivative: $f'(x) = \lim_{{h} \to 0} \frac{\sin(x+h) - \sin(x)}{h}$

Step 2: Utilizing the trigonometric identity: $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$ Plugging this in for $A = x$ and $B = h$, we get: $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$

Step 3: Substituting this expansion into our derivative definition, we get: $f'(x) = \lim_{{h} \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$

Step 4: Grouping terms: $f'(x) = \sin(x) \lim_{{h} \to 0} \frac{\cos(h) - 1}{h} + \cos(x) \lim_{{h} \to 0} \frac{\sin(h)}{h}$

Step 5: As $h \to 0$, the limit $\frac{\cos(h) - 1}{h}$ approaches 0, which can be derived using L'Hôpital's Rule or Taylor series. This will make the first term zero.

Step 6: For the second term, we recognize a very important limit: $\lim_{{h} \to 0} \frac{\sin(h)}{h} = 1$

To see this, consider the unit circle. If you take a small angle $h$ (measured in radians), the arc of the circle corresponding to that angle has a length approximately $h$. The vertical line segment from the x-axis to the arc has a length $\sin(h)$. As the angle $h$ approaches 0, the arc of the circle and the vertical line segment both become very small and approach the same length, making the ratio of their lengths approach 1.

Step 7: Plugging in this limit result: $f'(x) = 0 + \cos(x) = \cos(x)$

Thus, the derivative of $\sin(x)$ with respect to $x$ is $\cos(x)$.

Proof for the Derivative of $\cos(x)$ Being $-\sin(x)$

Using the proof we derived for the derivative of $\sin(x)$ being $\cos(x)$, we can deduce the derivative of $\cos(x)$.

Step 1: Start with the definition of the derivative: $f'(x) = \lim_{{h} \to 0} \frac{\cos(x+h) - \cos(x)}{h}$

Step 2: Utilizing the trigonometric identity: $\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ Plugging this in for $A = x$ and $B = h$, we get: $\cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h)$

Step 3: Substituting this expansion into our derivative definition, we have: $f'(x) = \lim_{{h} \to 0} \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h}$

Step 4: Grouping terms: $f'(x) = \cos(x) \lim_{{h} \to 0} \frac{\cos(h) - 1}{h} - \sin(x) \lim_{{h} \to 0} \frac{\sin(h)}{h}$

Step 5: From the proof of the derivative of $\sin(x)$, we have: $\lim_{{h} \to 0} \frac{\cos(h) - 1}{h} = 0$ and $\lim_{{h} \to 0} \frac{\sin(h)}{h} = 1$

Step 6: Plugging in these limit results: $f'(x) = 0 - \sin(x) = -\sin(x)$

Thus, the derivative of $\cos(x)$ with respect to $x$ is $-\sin(x)$.

By leveraging the previous proof for the derivative of $\sin(x)$, we're able to succinctly determine the derivative of $\cos(x)$.

Proof for the Derivative of $e^x$

Step 1: Start with the definition of the derivative:

$f'(x) = \lim_{{h} \to 0} \frac{e^{x+h} - e^x}{h}$

Step 2: Utilize properties of exponentials:

We know that $e^{x+h} = e^x \cdot e^h$. Using this property, we can rewrite the above as:

$f'(x) = \lim_{{h} \to 0} \frac{e^x \cdot e^h - e^x}{h}$

Step 3: Factor out the common term $e^x$:

$f'(x) = e^x \lim_{{h} \to 0} \frac{e^h - 1}{h}$

Step 4: Use the defining property of $e$:

By definition, the number $e$ is the unique positive number such that:

$\lim_{{h} \to 0} \frac{e^h - 1}{h} = 1$

This is essentially the definition of the number $e$. Using this property, the expression becomes:

$f'(x) = e^x \cdot 1 = e^x$

Therefore, the derivative of $e^x$ with respect to $x$ is $e^x$.

Derivative of $ \ln(x) $

Given:

The derivative of $e^x$ is $e^x$.

Goal:

To find the derivative of $\ln(x)$.

Proof:

Step 1: Define the function:

Let $y = \ln(x)$.

Step 2: Use properties of logarithms:

Using the inverse property of logarithms and exponentials, we can write:

$x = e^y$

Step 3: Differentiate both sides with respect to $x$:

Using the chain rule, the left side becomes \frac{dx}{dx} = 1 $ and the right side becomes $e^y \cdot \frac{dy}{dx}.

So,

$1 = e^y \cdot \frac{dy}{dx}$

Step 4: Solve for $ \frac{dy}{dx} $:

$\frac{dy}{dx} = \frac{1}{e^y}$

Step 5: Substitute back the original function:

From step 1, we know $y = \ln(x)$, so $e^y = x$.

Therefore,

$\frac{dy}{dx} = \frac{1}{x}$

The derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$.