91Ó°ÊÓ

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 \(1\cdot e^{x} = e^{x}\).

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 \(k e^{k x}\).