Derivative of 2xe^x

We now understand the derivative of 2xe^x. It is commonly represented as d/dx (2xe^x) or (2xe^x)', and its value is 2xe^x + 2e^x. The function 2xe^x has a well-defined derivative, indicating it is differentiable across its domain.

The key concepts are mentioned below:

Exponential Function: (e^x is the base of natural logarithms).

Product Rule: Rule for differentiating 2xe^x (since it consists of a product of 2x and e^x).

The derivative of 2xe^x can be denoted as d/dx (2xe^x) or (2xe^x)'. The formula we use to differentiate 2xe^x is: d/dx (2xe^x) = 2xe^x + 2e^x The formula applies to all x in the real number system.

We can derive the derivative of 2xe^x using proofs. To show this, we will use differentiation rules.

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

Using Product Rule

Verification by First Principle We will now demonstrate that the differentiation of 2xe^x results in 2xe^x + 2e^x using the above-mentioned methods:

Using Product Rule

To prove the differentiation of 2xe^x using the product rule, Consider f(x) = 2x and g(x) = e^x So we get, 2xe^x = f(x)·g(x) By product rule: d/dx [f(x)·g(x)] = f'(x)·g(x) + f(x)·g'(x) Let’s substitute f(x) = 2x and g(x) = e^x, d/dx (2xe^x) = (2)(e^x) + (2x)(e^x) = 2e^x + 2xe^x Thus, d/dx (2xe^x) = 2xe^x + 2e^x

Verification by First Principle

The derivative of 2xe^x can be verified using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 2xe^x using the first principle, we will consider f(x) = 2xe^x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [(f(x + h) - f(x)) / h] Given that f(x) = 2xe^x, we write f(x + h) = 2(x+h)e^(x+h). Substituting these into the equation, f'(x) = limₕ→₀ [(2(x + h)e^{(x + h)} - 2xe^x) / h] = limₕ→₀ [2xe^{(x + h)} + 2he^{(x + h)} - 2xe^x] / h = limₕ→₀ [2xe^{x(eh - 1)} + 2he^{(x + h)}] / h By dividing each term by h, we have: = limₕ→₀ [2xe^x (e^h - 1)/h + 2e^{(x + h)}] Using the limit formula limₕ→₀ (e^h - 1)/h = 1, = 2xe^x (1) + 2e^x = 2xe^x + 2e^x Hence, verified.

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a bit complex. To understand them better, consider a car whose speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like 2xe^x.

For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.

For the nth Derivative of 2xe^x, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.

When x = 0, the derivative of 2xe^x = 2xe^x + 2e^x = 0 + 2 = 2.

When x approaches negative infinity, the derivative tends toward 0, as e^x approaches 0.

• Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.

• Exponential Function: A mathematical function involving exponents, often represented as e^x in natural logarithms.

• Product Rule: A rule used to differentiate functions that are products of two or more functions.

• First Principle: A method to derive the derivative using the limit of the difference quotient.

• Rate of Change: A measure of how a quantity changes over time or in relation to another variable.