91Ó°ÊÓ

In calculus, the chain rule is a fundamental method for finding the derivative of composite functions. It helps in differentiating a function that is nested within another function. For instance, if you have a function where one variable is dependent on another, which is, in turn, dependent on a third variable, the chain rule becomes essential. Let's denote the inner function as $$u$$ and the outer function as $$f(u)$$. According to the chain rule, the derivative of $$f(g(x))$$ is:

\ \frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x) \

In this exercise, we need to differentiate $$e^{-k x}$$; here, the inner function is $$-kx$$, and the outer function is $$e^{u}$$ with $$u = -kx$$. Using the chain rule, the derivative becomes
$$ \frac{d}{dx}(e^{-k x}) = e^{-k x} \times (-k) $. This application of the chain rule simplifies the differentiation process, especially for complex functions.

Exponential functions often appear in calculus and involve terms like $$e^{u}$$ where $$e$$ is the base of natural logarithms, approximately equal to 2.718. The unique property of the exponential function is that it is its own derivative. Mathematically, this is represented as:

\ \frac{d}{du} e^u = e^u \ When an exponential function contains a more complex exponent, like $$u = -kx$$, we can use the chain rule to find the derivative effectively. For example, differentiating $$e^{-k x}$$ involves considering the inner function $$-kx$$ first, then applying the chain rule as shown before. The result is: $$ \frac{d}{dx}(e^{-kx}) = e^{-kx} \times (-k) $$. This property makes exponential functions particularly convenient when dealing with growth or decay problems.

Differentiating constant terms is one of the simplest operations in calculus. A constant term is a number on its own, without a variable attached to it. When you differentiate a constant, the result is always zero. This is because a constant does not change, so its rate of change or derivative is zero. Mathematically, if $$c$$ is a constant, then

\ \frac{d}{dx}(c) = 0 \

In our exercise, the function given is $$y = 1 - e^{-kx}$$. We identify the constant term here as $$1$$. Differentiating $$1$$ with respect to $$x$$, we get $$0$$. This understanding simplifies the differentiation of expressions involving constant terms.

After finding the derivatives of individual terms in a function, the next step is to combine them properly. This involves summing or subtracting the derivatives according to the original function's structure. In our function, $$y = 1 - e^{-kx}$$, we found that

\ \frac{d}{dx}(1) = 0 \ and \ \frac{d}{dx}(-e^{-kx}) = -e^{-kx} \times (-k) = ke^{-kx} \

To combine these results, we simply add the derivatives together following the original function's layout: $$ y' = 0 - (-ke^{-kx}) $$. This simplifies to $$ y' = ke^{-kx} $$. Combining derivatives correctly is crucial for obtaining the final result of the differentiation process.