Advanced Rules of Differentiation: Chain, Product, and Quotient Applications

Differentiation of Composite Functions using the Chain Rule

The process of differentiating a function that consists of a rational expression raised to a power involves the application of the Chain Rule in conjunction with the Quotient Rule. Consider the function:

$y = \left( \frac{3x-1}{2x+1} \right)^2$

To find $\frac{dy}{dx}$, we utilize a substitution method where we let the internal expression be represented by a variable $u$:

$\text{Let } u = \frac{3x-1}{2x+1}$

Consequently, the original function can be rewritten in terms of $u$ as:

$y = u^2$

Following the Chain Rule, the derivative of $y$ with respect to $x$ is given by the product of the derivative of $y$ with respect to $u$ and the derivative of $u$ with respect to $x$:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

First, we calculate $\frac{dy}{du}$ by differentiating $y = u^2$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(u^2) = 2u$

Substituting the original expression for $u$ back into this result, we get:

$\frac{dy}{du} = 2 \left( \frac{3x-1}{2x+1} \right)$

Next, we calculate $\frac{du}{dx}$ using the Quotient Rule, which is defined as $\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$. Applying this to $u = \frac{3x-1}{2x+1}$, where the numerator is $(3x-1)$ and the denominator is $(2x+1)$, we have:

$\frac{du}{dx} = \frac{(2x+1) \frac{d}{dx}(3x-1) - (3x-1) \frac{d}{dx}(2x+1)}{(2x+1)^2}$

Evaluating the derivatives in the numerator:

$\frac{du}{dx} = \frac{(2x+1)(3) - (3x-1)(2)}{(2x+1)^2}$

Expanding the terms:

$\frac{du}{dx} = \frac{6x+3 - (6x-2)}{(2x+1)^2} = \frac{6x+3-6x+2}{(2x+1)^2} = \frac{5}{(2x+1)^2}$

Finally, we combine the two components to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = 2 \left( \frac{3x-1}{2x+1} \right) \times \frac{5}{(2x+1)^2}$

$\frac{dy}{dx} = \frac{10(3x-1)}{(2x+1) (2x+1)^2} = \frac{10(3x-1)}{(2x+1)^3}$

The Product Rule for Differentiation

The Product Rule is used when a function is the product of two differentiable functions. The rule is stated as:

$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

Given the function:

$y = (x-1)(2x+5)$

To find $\frac{dy}{dx}$, we assign $u = (x-1)$ and $v = (2x+5)$. The derivative is calculated as follows:

$\frac{dy}{dx} = (x-1) \frac{d}{dx}(2x+5) + (2x+5) \frac{d}{dx}(x-1)$

Carrying out the differentiation for each term:

$\frac{dy}{dx} = (x-1)(2) + (2x+5)(1)$

Simplifying the expression by distributing the constants:

$\frac{dy}{dx} = 2(x-1) + 2x+5$

$\frac{dy}{dx} = 2x-2+2x+5 = 4x+3$

The Quotient Rule for Differentiation

The Quotient Rule is applied when differentiating the ratio of two functions. It is represented as:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Consider the function defined as:

$y = \frac{3x+4}{4x+5}$

Here, we let the numerator be $u = 3x+4$ and the denominator be $v = 4x+5$. To find $\frac{dy}{dx}$, we substitute these into the Quotient Rule formula:

$\frac{dy}{dx} = \frac{(4x+5) \frac{d}{dx}(3x+4) - (3x+4) \frac{d}{dx}(4x+5)}{(4x+5)^2}$

Evaluating the derivatives of the linear expressions:

$\frac{dy}{dx} = \frac{(4x+5)(3) - (3x+4)(4)}{(4x+5)^2}$

Expanding the numerator:

$\frac{dy}{dx} = \frac{12x + 15 - (12x + 16)}{(4x+5)^2}$

$\frac{dy}{dx} = – \frac{1}{(4x+5)^2}$

Differentiation of Root Functions containing Rational Expressions

When a function involves a square root of a quotient, such as:

$y = \sqrt{\frac{1+x}{1-x}}$

We determine $\frac{dy}{dx}$ by employing the Chain Rule. We start by letting the radicand be $u$:

$\text{Let } u = \frac{1+x}{1-x}$

Then the function becomes:

$y = \sqrt{u} = u^{\frac{1}{2}}$

Differentiating $y$ with respect to $u$ using the Power Rule:

$\frac{dy}{du} = \frac{d}{du}(u^{\frac{1}{2}}) = \frac{1}{2}u^{\frac{1}{2}-1} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}$

Now we differentiate the substitution $u = \frac{1+x}{1-x}$ with respect to $x$ using the Quotient Rule:

$\frac{du}{dx} = \frac{(1-x) \frac{d}{dx}(1+x) - (1+x) \frac{d}{dx}(1-x)}{(1-x)^2}$

Substituting the derivatives:

$\frac{du}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2}$

Simplifying the numerator terms:

$\frac{du}{dx} = \frac{1-x - (-1-x)}{(1-x)^2} = \frac{1-x+1+x}{(1-x)^2} = \frac{2}{(1-x)^2}$

Finally, we find $\frac{dy}{dx}$ by multiplying the results from both steps according to the Chain Rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{u}} \times \frac{2}{(1-x)^2}$

$\frac{dy}{dx} = \frac{1}{\sqrt{\frac{1+x}{1-x}}} \times \frac{1}{(1-x)^2}$

To simplify this further, we can rewrite the root of the fraction:

$\frac{1}{\sqrt{\frac{1+x}{1-x}}} = \frac{\sqrt{1-x}}{\sqrt{1+x}}$

Combining the expressions:

$\frac{dy}{dx} = \frac{\sqrt{1-x}}{\sqrt{1+x}(1-x)^2}$

Since $(1-x)^2 = \sqrt{1-x}^4$, we can cancel a factor of $\sqrt{1-x}$. Alternatively, using exponents, we have $(1-x)^{-2} \times (1-x)^{\frac{1}{2}} = (1-x)^{-\frac{3}{2}}$. This results in:

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