Question

Differentiate the following functions: \(\frac{\left(s^{2}-a^{2}\right)^{2}}{s^{3}}\)

Short answer

Question: Find the derivative of the function \(f(s) = \frac{(s^2 - a^2)^2}{s^3}\) Answer: The derivative of the given function is: \(f'(s) = \frac{s^2(s^2 - a^2)(4s^2 - 3(s^2 - a^2))}{s^6}\)

Step-by-step solution

Define g(s) and h(s) and find their derivatives

Let's define the functions g(s) and h(s) as follows: \(g(s) = (s^2 - a^2)^2\) \(h(s) = s^3\) Now, find the derivatives of these functions: Using the chain rule for g(s) and the power rule for h(s), \(g'(s) = 2(s^2 - a^2) \cdot 2s = 4s(s^2-a^2)\) \(h'(s) = 3s^2\)

Apply the quotient rule

Now that we have g'(s) and h'(s), we can apply the quotient rule to find the derivative of the given function: \(f'(s) = \frac{g'(s)h(s) - g(s)h'(s)}{h(s)^2}\) By substituting \(g'(s)\), \(h(s)\) and \(h'(s)\), we get: \(f'(s) = \frac{4s(s^2-a^2)(s^3) - (s^2 - a^2)^2(3s^2)}{s^6}\)

Simplify the expression

Finally, let's simplify the expression: \(f'(s) = \frac{4s^4(s^2-a^2) - 3s^2(s^2 - a^2)^2}{s^6}\) Factor out \(s^2(s^2-a^2)\) from the numerator: \(f'(s) = \frac{s^2(s^2 - a^2)(4s^2 - 3(s^2 - a^2))}{s^6}\) Now, the function has been differentiated and simplified.

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus that allows us to differentiate compositions of functions. When you have a function nested inside another, like a Russian doll, the chain rule helps you take it apart, derivative piece by derivative piece. It's expressed as \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

In our exercise, we used the chain rule to differentiate \( g(s) = (s^2 - a^2)^2 \). Normally, you'd be stuck pondering how to tackle the square of that binomial, but the chain rule guides you to first differentiate the outer function—imagine peeling an onion's outermost layer—then multiply it by the derivative of the inner function, peeling the next layer of the onion. It's integral (pun intended!) to keep things moving smoothly in calculus just like oil in an engine.

Power Rule

The power rule is the bread and butter of differentiating functions with exponents. It's beautifully simple: to differentiate \( x^n \), you bring down the exponent and multiply it by the base raised to one less than the original power to get \( nx^{n-1} \).

When we applied it to \( h(s) = s^3 \) in our problem, we simply took that exponent of three, brought it down with the confidence of a paratrooper jumping out of a plane, and multiplied by \( s \) squared. The power rule is what you'd call the 'calculus Swiss Army knife' because it comes in handy in so many situations—especially paired with other rules, like in our initial problem.

Quotient Rule

Ever encountered a fraction that you needed to differentiate? That's where the quotient rule swoops in. This rule is all about dealing with the division of two functions. It's a bit more complex, with a catchy rhythm to remember it: 'low dee high minus high dee low, square the bottom and away we go.' In latex terms, for two functions \( g(x) \) and \( h(x) \), the quotient rule is \( (\frac{g(x)}{h(x)})' = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \).

Our exercise example asked us to call this rule into action to tame the division within \( \frac{(s^2-a^2)^2}{s^3} \). It's all about following the formula with precision—just like following a recipe to bake a victorious cake—mix the functions and their derivatives in the right order, and voila, the derivative is served.

Calculus Derivatives

Calculus derivatives can seem daunting, but they're essential in measuring change. Think of them as the mathematical equivalent of a microscope. Derivatives allow us to zoom in on a graph and find out exactly how steep (or not) it is at any given point. With derivatives, we can tell at a glance if a function is growing faster than a weed in your garden or if it's as stagnant as a pond with no outlet.

They encapsulate the rate at which things move and change, from the orbits of planets to how fast a cup of coffee cools off. Derivatives are everywhere, and understanding how to use various rules like the chain, power, and quotient rules equips you with a toolkit to tackle all differentiable functions—just a fancy name for functions you can zoom in on indefinitely.