Euler’s Number and Derivatives Revision Notes for HSC SSCE Mathematics Advanced

Euler's Number and Derivatives

What is Euler's number?

Euler's number is a special mathematical constant denoted by the letter $e$ . It is an irrational number, which means it cannot be expressed as a simple fraction and its decimal representation goes on forever without repeating.

The approximate value of Euler's number is:

$e \approx 2.71828182845...$

What makes Euler's number truly remarkable is its unique relationship with differentiation. It is the only base for an exponential function where the derivative of the function equals the function itself.

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Euler's number is named after the Swiss mathematician Leonhard Euler. While other exponential functions like $2^x$ or $10^x$ change their behavior when differentiated, only e^x remains unchanged through the differentiation process.

The derivative of $e^x$

The exponential function $y = e^x$ has an extraordinary property:

$\frac{d}{dx}(e^x) = e^x$

This means that when we differentiate $e^x$ , we simply get $e^x$ back. The function is its own derivative.

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What does this mean? The rate of change of the function at any point equals the value of the function at that point. If you're standing on the curve $y = e^x$ at some point, the slope of the tangent line at that point equals the $y$ -coordinate where you're standing.

This property distinguishes $e^x$ from all other exponential functions. For example, if we consider $y = a^x$ where $a$ is any positive number other than $e$ , then $\frac{d}{dx}(a^x) = a^x$ only when $a = e$ .

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Worked Example: Finding the derivative at $x = 0$

Let's find the value of the derivative of $y = e^x$ at $x = 0$ .

Strategy: Use the property $\frac{dy}{dx} = e^x$ and evaluate at $x = 0$ .

Solution:

$\frac{dy}{dx} = e^x$

Substituting $x = 0$ :

$\frac{dy}{dx} = e^0 = 1$

The gradient at x = 0 is 1, confirming that $\frac{dy}{dx} = e^x$ .

Note: This result tells us that at the point $(0, 1)$ on the curve $y = e^x$ , the tangent line has a slope of $1$ .

Tangent to exponential functions

Because we know that the derivative of $y = e^x$ is $e^x$ , we can find the equation of the tangent line at any point on the curve. The derivative gives us the gradient at any point $(x, e^x)$ , which is simply $e^x$ .

To find the equation of a tangent line, we use the point-slope form:

$y - y_1 = m(x - x_1)$

where $m = e^x$ is the gradient and $(x_1, e^{x_1})$ is the point of tangency.

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At any point on $y = e^x$ , the gradient of the tangent equals the $y$ -coordinate. For example, at the point $(1, e)$ , the gradient is $m = e$ .

The graph shows the curve $y = e^x$ passing through the points $(0, 1)$ and $(1, e)$ . The tangent line at the point $(1, e)$ has gradient $m = e$ .

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Worked Example: Finding the tangent equation at $x = 1$

Let's determine the equation of the tangent to $y = e^x$ at $x = 1$ .

Strategy: Find the point and gradient at $x = 1$ using $\frac{d}{dx}(e^x) = e^x$ , then use point-slope form.

Solution:

First, find the point: At $x = 1$ , $y = e^1 = e$ , so the point is $(1, e)$ .

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

The gradient at x = 1 is e.

Now use the point-slope form:

y - e = e(x - 1) y = e(x - 1) + e y = ex - e + e y = ex

The equation of the tangent is y = ex.

Reflect and check: We can verify this by substituting $x = 1$ into $y = ex$ : $y = e \times 1 = e$ , which matches the point $(1, e)$ .

Important terminology

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Derivative: The result obtained after differentiation. For a function $f(x)$ , the derivative is the gradient function of $f(x)$ and is denoted $f'(x)$ .

For the exponential function $y = e^x$ , the derivative is $e^x$ at any point.

Remember!

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Key Points to Remember:

Euler's number $e \approx 2.71828182845$ is a special irrational constant
The exponential function $y = e^x$ has the unique property that $\frac{d}{dx}(e^x) = e^x$ - it equals itself when differentiated
This means the slope at any point on the curve $y = e^x$ equals the $y$ -coordinate at that point
To find the equation of a tangent to $y = e^x$ at point $(x, e^x)$ , use point-slope form with gradient $m = e^x$
At $x = 0$ , the derivative is $1$ (since $e^0 = 1$ ), and at $x = 1$ , the derivative is $e$ (since $e^1 = e$ )

Euler’s Number and Derivatives (HSC SSCE Mathematics Advanced): Revision Notes