Problem 12 - Calculus an Applied Approach

Question

Find \(d y / d u, d u / d x,\) and \(d y / d x.\) $$ y=2 \sqrt{u}, u=5 x+9 $$

Step-by-Step Solution

Answer

The derivative of \(y\) with respect to \(u\) is \( u^{-1/2}\), the derivative of \(u\) with respect to \(x\) is 5, and the derivative \(y\) with respect to \(x\) is \(5u^{-1/2}\).

Step 1: Find \(d y / d u\)

Take the derivative of \(y = 2 \sqrt{u}\) with respect to \(u\). Using the power rule, which states that the derivative of \(u^n\) is \(n*u^{n-1}\), where \(n\) is a real number, rewrite \(\sqrt{u}\) as \(u^{1/2}\) then apply the rule. The derivative is \(d y / d u = 2 * 1/2 * u^{-1/2} = u^{-1/2}\).

Step 2: Find \(d u / d x\)

Take the derivative of \(u = 5x + 9\) with respect to \(x\). The derivative is simply the coefficient of \(x\), since the derivative of a constant is 0. Thus, \(d u / d x = 5\).

Step 3: Find \(d y / d x\)

Finally, calculate \(d y / d x\) by multiplying \(d y / d u\) and \(d u / d x\). According to the chain rule, \(d y / d x = (d y / d u) * (d u / d x)\). Substituting the derivatives calculated in steps 1 and 2, we get \(d y / d x = u^{-1/2} * 5 = 5u^{-1/2}\).

Key Concepts

Derivative of a Function

Understanding the derivative of a function is crucial for diving into any calculus problem. In essence, the derivative represents the rate at which a function's output value changes as its input value changes. It's like capturing the 'instantaneous' change of the function at any point. In practical terms, if you were to graph the function, the derivative at any point would give you the slope of the tangent line to the function at that point.

For instance, in the exercise where we are given the function \(y = 2 \sqrt{u}\), finding the derivative \(dy/du\) gives us the rate at which \(y\) changes as \(u\) changes. In other words, if \(u\) were time and \(y\) were distance, \(dy/du\) would tell us the speed at a particular instant. Derivative calculations often use rules like the power rule, sum rule, and chain rule to simplify the process, making them more accessible for students to perform complex calculus operations like these.

Power Rule

The power rule is one of the most fundamental and frequently used rules in computing derivatives. It states that if you have a function \(u^n\), where \(n\) is a real number, its derivative with respect to \(u\) is \(n \cdot u^{n-1}\). The beauty of the power rule is in its simplicity and broad application; it can be applied whether \(n\) is positive, negative, or even a fraction.

In our specific exercise, the power rule is used to find the derivative of \(y = 2\sqrt{u}\) by rewriting the square root of \(u\) as \(u^{1/2}\) and then applying the power rule to find that \(dy/du = u^{-1/2}\). Understanding this rule deeply allows students to tackle a wide range of problems involving powers of variables. It's not just a rote calculation; it's a powerful tool that simplifies what could otherwise be very complex differentiation problems.

Implicit Differentiation

Implicit differentiation is a technique used when dealing with functions that are not explicitly solved for one variable in terms of another. This means you might have an equation where both \(x\) and \(y\) are mixed together, and you can't easily get a 'y=' or an 'x=' statement. However, instead of isolating variables which can be complex or sometimes impossible, you assume that both \(y\) and \(x\) are functions of a common variable (often \(t\), representing time) and differentiate both sides of the equation with respect to that variable.

Although our exercise doesn't require implicit differentiation since the functions are given explicitly, understanding this concept is essential for more complicated scenarios. This method is often utilized when applying the chain rule, which we've used to find the derivative \(dy/dx\) in this exercise. Implicit differentiation ensures that students are well-prepared for a variety of calculus problems, not only those with neatly arranged functions.