Question

Find the derivative of each function. Check some by calculator. $$y=\frac{1}{\sqrt{x^{2}-x}}$$

Short answer

\(\frac{dy}{dx} = \frac{1-x}{(x^2-x)^{\frac{3}{2}}}\)

Step-by-step solution

Identify the Function to Differentiate

First, write down the function to differentiate, which is given as \(y = \frac{1}{\sqrt{x^2 - x}}\).

Rewrite the Function

Rewrite the function in a form that is easier to differentiate, using the power rule. The function can be rewritten as \(y = (x^2 - x)^{-\frac{1}{2}}\).

Apply the Chain Rule

Use the chain rule to differentiate the function with respect to x. The chain rule states that the derivative of \(f(g(x))\), where f and g are functions, is \(f'(g(x))g'(x)\).

Differentiate Using the Power Rule

According to the power rule, \(\frac{d}{dx}x^n = nx^{n-1}\). So, differentiate the outer function using the power rule, which results in \(-\frac{1}{2}(x^2 - x)^{-\frac{3}{2}}\).

Differentiate the Inside Function

Now differentiate the inside function \(u = x^2 - x\), which yields \(u' = 2x - 1\).

Apply the Derivative of the Inside Function

Multiply the derivative of the outside function by the derivative of the inside function to complete the application of the chain rule. The final derivative of y with respect to x is \(\frac{dy}{dx} = -\frac{1}{2}(x^2 - x)^{-\frac{3}{2}}(2x - 1)\).

Simplify the Expression

You can simplify the expression to obtain a final answer for the derivative, which is \(\frac{dy}{dx} = -\frac{1}{2} \cdot \frac{2x - 1}{(x^2 - x)^{\frac{3}{2}}} = \frac{1-x}{(x^2-x)^{\frac{3}{2}}}\).

Key concepts

Chain Rule

In derivative calculus, the chain rule is a fundamental method for finding the derivative of composite functions. It's like a mathematical assembly line; you break down the function into its parts, find the derivatives individually, and then put it all back together.
Whenever you have a function within a function, the chain rule allows you to differentiate this multi-layered situation with ease. The magic formula here is: if you have a function defined as \( f(g(x)) \), the derivative, typically noted as \( f'(x) \) or \( \frac{df}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).
For a more tangible approach, think of the original function as a nesting doll; you take each doll apart, differentiate each layer, and then reassemble them. It's important to first identify the inner function (\( g(x) \) in our case) and the outer function (\( f(u) \) where \( u=g(x) \) ), before applying the derivatives respectively.

Power Rule

The power rule is another cornerstone of derivative calculus. It's the go-to tool for handling functions where the variable is raised to a power. The rule is straightforward: for \( x^n \), the derivative is \( nx^{n-1} \).
Consider it the bread and butter for simple differentiation. It's particularly helpful because of its simplicity and how frequently it applies. You can tackle a variety of functions with the power rule, as long as they have that clear exponent form. This is invaluable because, in the world of derivative calculus, a lot of complex-looking functions can often be simplified to a form where the power rule can be applied. Remember that constants (like 2 in \( 2x \) ) come along for the ride, multiplying the result.

Derivative Calculus

Derivative calculus is the process of finding the rate at which a function changes at any given point. It's the heartbeat of calculus, capturing the essence of change. When you work out the derivative, you're essentially finding the slope of the tangent line to a function at any point you choose.
Think of it as a snapshot of a rocket's speed at a specific moment during its launch. You're not just dealing with the journey; you're zeroing in on a single, pivotal moment in time. Derivative calculus is applied in various fields from physics to economics because understanding the rate of change is universally important.

Function Simplification

Function simplification in calculus is about making complex expressions more manageable. Simplifying a function can help you identify its behavior, zeroes, and integrals, or even just make it easier to draw. When differentiating, your first battle is often with the algebra: factor, cancel, and reduce whenever you can to make your differentiation straightforward.
An expert tip: keep an eye out for opportunities to use identities or common patterns to simplify before you dive into calculus. It can turn a daunting derivative into a walk in the park. Simplification can be applied at any stage of solving, whether before you differentiate (as with rewriting \( y = \frac{1}{\sqrt{x^2 - x}} \) in our example), or after you've taken the derivative, to tidy up your final answer.