Rivalry2

HTML Diff

0 added 0 removed

Original 2026-01-01

Modified 2026-02-28

1 We can derive the derivative of \(e^x\) using proofs.

2 To show this, we will use the properties of exponential functions along with the rules of differentiation.

3 There are several methods we use to prove this, such as:

4 By First Principle

5 Using Chain Rule

6 Using Product Rule

7 We will now demonstrate that the differentiation of \(e^x\) results in e^x using the above-mentioned methods:

8 By First Principle

9 The derivative of \(e^x\) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.

10 To find the derivative of \(e^x\) using the first principle, we will consider f(x) = \(e^x\).

11 Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

12 Given that f(x) = \(e^x\), we write f(x + h) = \(e^(x + h)\).

13 Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [\(e^(x + h) - e^x\)] / h = limₕ→₀ [\(e^x · (e^h - 1)\)] / h = \(e^x \)· limₕ→₀ [\((e^h - 1)\) / h]

14 Using the limit property, limₕ→₀ (\(e^h - 1\)) / h = 1. f'(x) = \(e^x\) · 1 = \(e^x\)

15 Hence, proved.

16 Using Chain Rule

17 To prove the differentiation of \(e^x \)using the chain rule,

18 We use the formula: If y = \(e^(u)\) where u = x, then dy/dx = dy/du · du/dx = \(e^u\) · 1 = \(e^x\)

19 Using Product Rule

20 We will now prove the derivative of \(e^x\) using the<a>product</a>rule.

21 The step-by-step process is demonstrated below:

22 Here, we consider the function as a product of two identical exponential functions: \(e^x · e^0 = e^x · 1\)

23 Using the product rule formula: d/dx [u·v] = u'·v + u·v' u = \(e^x\) and v = \(e^0\) u' = d/dx (\(e^x\)) = \(e^x \)v' = d/dx (\(e^0\)) = 0 d/dx (\(e^x\)) = \(e^x · 1 + e^x · 0 = e^x\)

24 Thus, d/dx (\(e^x\)) =\( e^x\).