Question

Find \(\mathrm{f}^{\prime}(2)\) if \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{g}(\mathrm{x})}\) and \(\mathrm{g}(\mathrm{x})=\int_{2}^{x} \frac{\mathrm{t}}{1+\mathrm{t}^{4}} \mathrm{dt}\).

Short answer

Answer: The value of f'(2) is \(\frac{2}{17}\).

Step-by-step solution

Apply Chain Rule with f(g(x)) formula

Use the chain rule to differentiate the f(x) = e^(g(x)): (f(g(x)))' = f'(g(x)) * g'(x).

Find f'(g(x))

Find the derivative of e^(g(x)) with respect to g(x): f'(g(x)) = d(e^(g(x)))/dg = e^(g(x)) (using the basic rule of derivatives for e^(u)).

Find g'(x) using the Fundamental Theorem of Calculus

Since g(x) = ∫(t/(1 + t^4) dt) from 2 to x, the derivative of g(x) with respect to x is simply the integrand: g'(x) = t/(1 + t^4).

Substitute g'(x) value in Chain Rule formula

Now substitute g'(x) in the chain rule formula: f'(x) = f'(g(x)) * g'(x) = e^(g(x)) * (t/(1 + t^4)).

Find f'(2)

Substitute x = 2 in f'(x) and find the value of f'(2): f'(2) = e^(g(2)) * (2/(1 + 2^4)). We already know that g(2) = ∫(t/(1 + t^4) dt) from 2 to 2, which is zero (integral from equal bounds), so g(2) = 0. Now, substituting this into f'(2): f'(2) = e^0 * (2/(1 + 16)) = 1 * (2/17) = 2/17. Therefore, f'(2) = \(\frac{2}{17}\).

Key concepts

The Chain Rule in Calculus

Understanding the chain rule is vital for calculus students. It's the door to effectively handling complex derivatives. Imagine you've dressed a function in multiple layers and you want to find its rate of change - that's when the chain rule comes into play. Like peeling an onion, the chain rule helps us differentiate composite functions, those consisting of functions within functions.

Our textbook exercise demonstrates a classic application of the chain rule. The derivative of the outer function is first taken, leaving the inner function untouched, then multiplied by the derivative of the inner function itself. It’s like a mathematical dance: every function gets its turn. This rule is indispensable, especially when dealing with real-world scenarios that involve compound changes, like acceleration of a car as it follows curved paths.

Exponential Function Derivatives

When it comes to exponential function derivatives, things are surprisingly straightforward, especially for functions of the form \(e^{u(x)}\), where \(u(x)\) is another function. It's pretty elegant: the derivative of \(e^{u(x)}\) with respect to \(x\) is just \(e^{u(x)}\) times the derivative of \(u(x)\). This is an exceptional case in the derivative world because the function \(e^x\) is its own derivative.

The reasons lie in the function’s natural growth properties - it models phenomena that grow at a rate proportional to their size. It's like looking at a bank account that earns interest continuously; the maths mirrors the exponential growth of your savings over time. In calculus, it's critical to get comfortable with \(e\), as it often represents growth and decay in real-world contexts.

Integral Calculus

Venturing into the realm of integral calculus can feel like a switch from sprinting to marathon running. Here, we're focusing on cumulative change over intervals. Instead of instantaneous rates of change, as in derivatives, integrals look at the total accumulation.

The Fundamental Theorem of Calculus is the keystone. It's breathtakingly simple and yet so powerful – it connects differentiation and integration, the two main operations in calculus. It tells us that we can find the total accumulated change by looking at the area under the rate-of-change function. While differentiation chops up the function into infinitesimally small pieces to study it closely, integration sweeps up the aggregate effect to get a big picture view.

And precisely this theorem was employed in our exercise to find the derivative of the function \(g(x)\), itself defined as an integral from 2 to \(x\). It's like the calculus version of a loop-the-loop – beautifully circling back on itself, showing the layered connections between these two fundamental concepts.