Question

Calculate the derivative of the following functions. $$y=\sqrt[4]{\frac{2 x}{4 x-3}}$$

Short answer

Answer: The derivative of the function is $$\frac{dy}{dx} = \frac{1}{4}\left(\frac{2x}{4x-3}\right)^{-\frac{3}{4}} \cdot \frac{(2)(4x-3) - (2x)(4)}{(4x-3)^2}$$

Step-by-step solution

Rewrite the function in a simpler form

Rewrite the given function as: $$y = (\frac{2x}{4x-3})^{\frac{1}{4}}$$ This will make it easier to apply the chain rule and the quotient rule.

Apply the chain rule

Let $$u = \frac{2x}{4x-3}$$, then $$y = u^{\frac{1}{4}}$$. We need to find $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$. First, find $$\frac{dy}{du}$$: $$\frac{dy}{du} = \frac{1}{4}u^{-\frac{3}{4}}$$

Apply the quotient rule to find du/dx

Now, find $$\frac{du}{dx}$$ using the quotient rule, which states: $$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$ Let $$f(x) = 2x$$ and $$g(x) = 4x - 3$$. Then, $$f'(x) = 2$$ and $$g'(x) = 4$$ So, $$\frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} = \frac{(2)(4x-3) - (2x)(4)}{(4x-3)^2}$$

Combine dy/dx

Now, find $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$: $$\frac{dy}{dx} = \frac{1}{4}u^{-\frac{3}{4}} \cdot \frac{(2)(4x-3) - (2x)(4)}{(4x-3)^2}$$

Substitute u back in and simplify the expression

Substitute $$u = \frac{2x}{4x-3}$$ back into the expression and simplify: $$\frac{dy}{dx} = \frac{1}{4}\left(\frac{2x}{4x-3}\right)^{-\frac{3}{4}} \cdot \frac{(2)(4x-3) - (2x)(4)}{(4x-3)^2}$$ You can further simplify the expression, but this is the derivative of the given function.

Key concepts

Chain Rule

The chain rule is a fundamental technique used in calculus to differentiate composite functions. It helps you find the derivative of a function by breaking it down into simpler parts.
For a given function, like in our example, you might have a composition of functions such as \( y = u^{\frac{1}{4}} \) where \( u = \frac{2x}{4x-3} \).
The chain rule expresses the derivative \( \frac{dy}{dx} \) through the formula:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

This means you first find the derivative of \( y \) with respect to \( u \) (\( \frac{dy}{du} \)), and then multiply it with the derivative of \( u \) with respect to \( x \) (\( \frac{du}{dx} \)).
This step-by-step approach is essential for simplifying problems involving nested functions.

Quotient Rule

The quotient rule is indispensable when dealing with the derivatives of functions expressed as fractions of two other functions. It is specifically used when you have a function in the form \( \frac{f(x)}{g(x)} \).
The formula for the quotient rule is as follows:

• \( \frac{d}{dx}(\frac{f(x)}{g(x)}) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \)

It's crucial to remember: the derivative of the top function \( f(x) \) is multiplied by the bottom function \( g(x) \), then subtract the product of the top function \( f(x) \) and the derivative of the bottom function \( g(x) \).
Finally, you divide the entire expression by the square of the bottom function, \( [g(x)]^2 \).
This enables you to differentiate functions that are structured as a quotient.

Simplification

Simplification plays a key role in solving complex mathematical derivations. By reworking expressions, you make them easier to differentiate or manipulate.
In our problem, the function \( y = \sqrt[4]{\frac{2 x}{4 x-3}} \) is rewritten upfront as \( y = (\frac{2x}{4x-3})^{\frac{1}{4}} \).
This transformation simplifies applying the chain rule and differentiating with respect to the variable.
Another part of simplification occurs after finding the derivative, where expressions can further be reduced for clarity or simplicity.
Simplification not only aids in calculation but also in verifying the correctness of derivatives.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function is changing at any point.
It's a foundational concept in calculus, providing insights into the behavior of functions.
In our exercise, differentiation involves several steps: applying the chain rule and quotient rule to find \( \frac{dy}{dx} \).
Ultimately, these steps let us express how \( y \) changes with respect to changes in \( x \).

• The result, \( \frac{dy}{dx} \), gives a precise measure of that change.

Differentiation provides powerful tools for understanding and solving real-world problems, from simple slopes to complex rate calculations.