Question

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x},\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this assumption implies that the rule is true for \(k=n+1\). (Hint: Write \(e^{(n+1) x}\) as the product of two functions and use the Product Rule.)

Short answer

Question: Prove by induction that the derivative of the function \(e^{k x}\), where k is a positive integer, is \(ke^{k x}\). Solution: We have proved by induction that the derivative of the function \(e^{k x}\) for positive integers \(k\) is \(k e^{k x}\).

Step-by-step solution

Part a: Base case (k=1)

To prove the statement for \(k=1\), we simply take the derivative of the function with respect to x. The function is \(e^{x}\). Differentiating with respect to x, we have: \(\frac{d}{d x}\left(e^{x}\right) = e^{x}\) From this, we can see that the statement is true for \(k=1\), since \(1\cdot e^{x} = e^{x}\).

Part b: Inductive step (prove for k=n+1, given k=n)

We assume the statement is true for \(k=n\). That is, \(\frac{d}{d x}\left(e^{n x}\right) = n e^{n x}\) Now we need to prove the statement for \(k=n+1\). Thus, we consider the function \(e^{(n+1) x}\). We can rewrite the function as the product of two functions: \(e^{(n+1) x} = e^{n x} \cdot e^{x}\) Now, we use the product rule to differentiate this expression with respect to x: \(\frac{d}{d x}\left(e^{(n+1) x}\right) = \frac{d}{d x}\left(e^{n x} \cdot e^{x}\right)\) The product rule states that the derivative of a product of two functions is: \(\frac{d}{d x}\left(u \cdot v\right) = u'v+uv'\) In this case, let \(u=e^{n x}\) and \(v=e^{x}\). Then, \(\frac{d}{d x}\left(u \cdot v\right) = \frac{d}{d x}\left(e^{n x} \cdot e^{x}\right) = \left(\frac{d}{d x} e^{n x}\right)e^{x} + e^{n x}\left(\frac{d}{d x} e^{x}\right)\) Recall that we assumed the statement is true for \(k=n\): \(\frac{d}{d x}\left(e^{n x}\right) = n e^{n x}\) And from our base case, we know that the derivative of \(e^{x}\): \(\frac{d}{d x}\left(e^{x}\right) = e^{x}\) Plugging this back into our equation, we get: \(\frac{d}{d x}\left(e^{(n+1) x}\right) = (n e^{n x})e^{x} + e^{n x}(e^{x}) = e^{n x}(n e^{x} + e^{x})\) Now, we factor out \(e^{x}\): \(\frac{d}{d x}\left(e^{(n+1) x}\right) = e^{n x}(e^{x}(n+1))\) This simplifies to: \(\frac{d}{d x}\left(e^{(n+1) x}\right) = (n+1)e^{(n+1) x}\) We have now proved the statement for \(k=n+1\). Thus, by induction, the derivative of \(e^{k x}\) for positive integers \(k\) is \(k e^{k x}\).

Key concepts

Derivative of Exponential Functions

Understanding the derivative of exponential functions is essential in calculus as they often occur in various scientific contexts, such as growth and decay models. An exponential function is of the form f(x) = e^{kx}, where e is the base of the natural logarithm, and k is a constant. The derivative describes the rate of change of a function. For exponential functions, an intriguing property is that the rate of change is proportional to the function's value.

When differentiating e^{kx}, the result is ke^{kx}. This stems from the fact that the base e has a unique property where its derivative is equal to itself, and multiplying by k accounts for the chain rule when kx is considered as a function inside the exponential function. This concept is foundational and appears frequently in solving both theoretical and practical problems in calculus.

Product Rule in Calculus

The product rule is a differentiation technique used when taking the derivative of a product of two functions. It is expressed as (uv)' = u'v + uv', where u and v are functions of x, and ' denotes the derivative with respect to x.

Understanding the Product Rule

Consider two smoothly varying quantities whose rates of change we wish to understand when they are multiplied together. Intuitively, the product's change reflects both items' independent changes and their combined effect. In calculus, this intuition is formalized by the product rule which allows us to calculate complex derivatives that involve multiplication of distinct functions effectively. It is an indispensable tool when dealing with exponential growth models, involving products of functions, or in physics, where it might represent forces acting on varying areas.

Differentiation Techniques

Differentiation techniques are various methods used to find the derivative of functions. These techniques range from basic rules, like the power rule and the constant multiple rule, to more advanced methods such as the chain rule and integration by parts.

Specifically, the derivatives of polynomial, trigonometric, exponential, and logarithmic functions are discovered using strategies customized to each function type.

• The power rule, for instance, handles polynomial functions
• Trigonometric functions often involve trigonometric identities alongside differentiation rules
• Exponential and logarithmic functions utilize their unique properties that simplify their derivatives

Mastering these techniques is crucial for students as they provide a toolkit to solve a broad spectrum of problems in calculus. Proper understanding ensures that students can tactically deconstruct a complex function into simpler parts that can be differentiated individually using the appropriate rules.