Question

Find \(d y / d u, d u / d x\), and \(d y / d x\). $$ y=u^{3}, u=3 x^{2}-2 $$

Short answer

After doing all the analysis and step-by-step calculations, the derivatives are: \(dy/du = 3u^{2}\), \(du/dx = 6x\), and \(dy/dx = 3(3x^{2}-2)^{2}*6x\).

Step-by-step solution

Finding \(dy/du\)

First off, start with the derivative of `y` with respect to `u`. The given function is \(y = u^3\). By using the power rule for differentiation, which states that the derivative of \(u^n\) is \(n u^{n-1}\), the derivative \(dy/du = 3 u^{2}\).

Finding \(du/dx\)

Next, find the derivative of `u` with respect to `x`. The given function is \(u = 3x^{2} - 2\). Apply the power rule for differentiation again and get \(du/dx = 6x\).

Finding \(dy/dx\)

Finally, find the derivative of `y` with respect to `x` by using chain rule. This involves multiplying the derivative of `y` with respect to `u`, and `u` with respect to `x`., which is \(dy/dx = dy/du * du/dx = 3u^{2}*6x\). However, since `u` is a function of `x`, i.e., \(u= 3x^{2} -2\), substitute `u` to get \(dy/dx = 3(3x^{2}-2)^{2}*6x\).

Key concepts

Power Rule

When you're learning about taking derivatives in calculus, the power rule is one of the simplest and most commonly used rules. It's a handy shortcut and very often your go-to method when dealing with polynomial expressions. The power rule states that for any function of the form \(u^n\), the derivative with respect to \(u\), denoted as \( \frac{d}{du}(u^n) \), is \(nu^{n-1}\). This is remarkably easy as you just bring down the exponent as a multiplier, and reduce the exponent by one.

In our problem, we needed to find \( \frac{dy}{du} \) where \(y = u^3\). By applying the power rule, this becomes \(3u^{2}\). Here is what happens step-by-step:

• Identify the power: in this case, it's 3.
• Take down the power as a coefficient: so the 3 from \(u^3\) moves in front.
• Decrease the original power by 1: moving from \(u^3\) to \(u^2\).

That's pretty much it! The power rule is as simple as it sounds and becomes second nature after some practice.

Chain Rule

The chain rule is an essential tool in calculus for dealing with composite functions. A composite function is simply a function nested inside another function, or to think of it as a "chain" of functions. The chain rule basically allows us to differentiate these kinds of functions easily, even when directly applying simpler rules like the power rule isn't enough.

In the exercise, we needed to find \( \frac{dy}{dx} \), and the relationship \(y = u^3\) and \(u = 3x^2 - 2\) gives rise to a composite situation. The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). In our case, we had already calculated:

• \( \frac{dy}{du} = 3u^{2} \)
• \( \frac{du}{dx} = 6x \)

We then simply multiply these together to apply the chain rule: \( \frac{dy}{dx} = (3u^{2})(6x) \), which becomes \(18xu^{2}\).

Finally, substitute \(u = 3x^2 - 2\) back into our equation to get \( \frac{dy}{dx} = 18x(3x^2 - 2)^2 \). This substitution step is crucial because you need everything in terms of \(x\) for a complete derivative.

Differentiation

Differentiation is the process of finding the derivative of a function, which essentially represents how a function changes. In simpler terms, it tells you the rate at which a quantity changes with respect to another quantity. Differentiation is the cornerstone of calculus and applies to almost any function you encounter.

In our exercise, differentiation allowed us to understand how changes in \(x\) affect \(u\) and consequently affect \(y\). We tackled this by finding separate derivatives:

• First, \( \frac{dy}{du} \), which showed how fast \(y\) changes as \(u\) changes.
• Second, \( \frac{du}{dx} \), which showed how fast \(u\) changes as \(x\) changes.
• Finally, combining them via the chain rule allowed us to find \( \frac{dy}{dx} \), which shows how fast \(y\) changes directly with \(x\).

Differentiation is far more than just an academic exercise, as it translates into real-world applications like measuring speed, understanding growth trends, and optimizing products or services. Understanding these fundamental exercises can empower you to solve many practical problems efficiently.