Question

Combining rules Compute the derivative of the following functions. $$h(x)=\frac{(x-1)\left(2 x^{2}-1\right)}{x^{3}-1}$$

Short answer

Question: Find the derivative of the function \(h(x)=\frac{(x-1)\left(2 x^{2}-1\right)}{x^{3}-1}\). Answer: The derivative of the function is \(h'(x) = \frac{-6x^4 + 7x^3 - 2x^2 - 3x + 2}{(x^3-1)^2}\).

Step-by-step solution

Identify the elements in the given function

We are given the function: $$h(x)=\frac{(x-1)\left(2 x^{2}-1\right)}{x^{3}-1}$$ Here, \(u(x)=x-1\), \(v(x)=2x^2-1\), and \(w(x)=x^3-1\). Rewriting our function as the quotient of two functions: $$h(x)=\frac{u(x)v(x)}{w(x)}$$

Calculate the derivatives of u, v, and w

Now, we will find the derivatives of u, v, and w with respect to x. For \(u(x)=x-1\): $$\frac{du}{dx} = 1$$ For \(v(x)=2x^2-1\): $$\frac{dv}{dx} = 4x$$ For \(w(x)=x^3-1\): $$\frac{dw}{dx} = 3x^2$$

Apply the Quotient Rule to find the derivative of h(x)

Now, using the quotient rule to find the derivative of h(x): $$h'(x) = \frac{w(x)\frac{d(uv)}{dx} - u(x)v(x)\frac{dw}{dx}}{w^2(x)}$$ We need to compute \(\frac{d(uv)}{dx}\) using the product rule.

Apply the Product Rule to find the derivative of the numerator (uv)

Using the product rule: $$\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$ Plugging in the previously calculated derivatives and original functions, we get: $$\frac{d(uv)}{dx} = (x-1)(4x) + (2x^2-1)(1)$$ Simplifying, we obtain: $$\frac{d(uv)}{dx} = 4x^2 - 4x + 2x^2 - 1$$

Plug everything back into the Quotient Rule

Now, let's replace all the expressions back into the quotient rule formula to find h'(x): $$h'(x) = \frac{(x^3-1)(4x^2-4x+2x^2-1)-(x-1)(2x^2-1)(3x^2)}{(x^3-1)^2}$$

Simplify the expression

Now, we will simplify the expression for h'(x): First, expand the numerator and combine like terms: $$h'(x)=\frac{(4x^2-4x+2x^2-1)(x^3-1)-(3x^2)(x-1)(2x^2-1)}{(x^3-1)^2}$$ $$h'(x)=\frac{6x^2x^3 - 6x^2 + x^3 - 1 - 6x^4 + 3x^2 + 6x^3 - 3x}{(x^3-1)^2}$$ Combining the like terms and simplifying the expression: $$h'(x) = \frac{-6x^4 + 7x^3 - 2x^2 - 3x + 2}{(x^3-1)^2}$$ So the derivative of the given function is: $$h'(x) = \frac{-6x^4 + 7x^3 - 2x^2 - 3x + 2}{(x^3-1)^2}$$

Key concepts

Quotient Rule

The quotient rule is a powerful tool in calculus used to find the derivative of a function that is the ratio of two differentiable functions. In our given function
\(h(x)=\frac{u(x)v(x)}{w(x)}\)
, where
\(u(x)\) and \(v(x)\) are the numerator functions and \(w(x)\) is the denominator function, the quotient rule is essential for dealing with this complexity. The formula for the quotient rule is:
\[h'(x) = \frac{w(x)\frac{d(uv)}{dx} - u(x)v(x)\frac{dw}{dx}}{w^2(x)}\].
In simpler terms, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. This formula might seem daunting at first, but with practice, it becomes a natural step in solving calculus problems involving division of functions.

Product Rule

When differentiating products of functions, the product rule comes into play. This rule states that to find the derivative of a product of two functions, you must multiply the first function by the derivative of the second and add it to the product of the second function and the derivative of the first. Mathematically, it's expressed as:
\[\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\].
It allows us to break down more complex expressions into simpler parts that can be differentiated individually. In the exercise, for instance, we used the product rule to differentiate the numerator \(u(x)v(x)\), resulting in \(4x^2 - 4x + 2x^2 - 1\). Proper application of the product rule is fundamental in solving calculus problems where functions are multiplied by one another.

Differentiation

Differentiation is at the core of calculus and is the process used to determine the rate at which a function is changing at any given point. For functions like \(u(x) = x - 1\), \(v(x) = 2x^2 - 1\), and \(w(x) = x^3 - 1\), finding the derivative is relatively straight-forward by applying basic rules of differentiation. For example, differentiating \(u(x)\) gives us a constant \(1\), since the derivative of a constant term is zero, and the derivative of \(x\) with respect to \(x\) is \(1\). The process involves applying known derivative rules such as the power rule, sum rule, and others to find the rate of change of each part of the function. Understanding how to apply these rules correctly is key to tackling a wide range of calculus problems.

Calculus Problems

Calculus problems often require combining the use of several rules and concepts to find derivatives, integrals, or solve for unknowns. The problems can range from basic to complex, requiring an understanding of fundamental principles such as limits, derivatives, integrals, and the theorems that connect them. In the case of the exercise provided, it was necessary to use both the product and quotient rules to find the derivative of the function \(h(x)\). The step-by-step solution broke down the process into identifiable parts that make it easier to follow and understand. Working through such problems enhances problem-solving skills and comprehension of calculus concepts, preparing students to tackle a wide variety of challenges they may encounter in mathematics and related fields.