2.2 The First Derivative
Differentiation is the algebraic process of finding the gradient function, \(f'(x)\) or \(\dfrac{dy}{dx}\), from a given function \(y=f(x)\). Instead of using the lengthy method of first principles, standard algebraic formulas and differentiation rules allow us to quickly determine the rate of change and the gradient of a curve at any specific point.
Learning Objectives
- Determine the first derivative of an algebraic function of the form \(y=ax^n\), where \(a\) is a constant and \(n\) is an integer or rational number.
- Use different notations to represent the first derivative: \(\dfrac{dy}{dx}\), \(f'(x)\), and \(\dfrac{d}{dx}[f(x)]\).
- Find the gradient of the tangent to a curve at a given point using its gradient function.
- Differentiate functions that contain terms algebraically added or subtracted term-by-term.
- Differentiate composite functions using the chain rule, and products and quotients using the product and quotient rules.
Summary of Differentiation Rules for Algebraic Functions
A quick-reference comparison of the basic power rule, chain rule, product rule and quotient rule, with cautionary notes about common differentiation errors.
1. Power Rule
\(y=ax^n \implies \dfrac{dy}{dx}=anx^{n-1}\)
2. Chain Rule
\(y=[u(x)]^n \implies \dfrac{dy}{dx}=n[u(x)]^{n-1}\cdot u'(x)\)
3. Product Rule
\(y=uv \implies \dfrac{dy}{dx}=uv'+vu'\)
4. Quotient Rule
\(y=\dfrac{u}{v} \implies \dfrac{dy}{dx}=\dfrac{vu'-uv'}{v^2}\)
| Rule | Formula | Common error to avoid |
|---|---|---|
| Power rule | \(\dfrac{d}{dx}(ax^n)=anx^{n-1}\) | Forgetting to multiply by the original power \(n\). |
| Chain rule | \(\dfrac{d}{dx}(u^n)=nu^{n-1}\dfrac{du}{dx}\) | Forgetting the inner derivative \(\dfrac{du}{dx}\). |
| Product rule | \(\dfrac{d}{dx}(uv)=uv'+vu'\) | Writing \(u'v'\) instead of \(uv'+vu'\). |
| Quotient rule | \(\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{vu'-uv'}{v^2}\) | Reversing the numerator order or writing \(\dfrac{u'}{v'}\). |
Short Explanation
Differentiating an algebraic function means finding a new function that outputs the gradient of the original graph at any value of \(x\).
1. First Derivative Formula for \(y=ax^n\)
If \(y=ax^n\), where \(a\) is a constant and \(n\) is an integer or real number, then \(\dfrac{dy}{dx}=anx^{n-1}\).
2. Notations for the First Derivative
If \(y=f(x)\), the first derivative can be written as \(\dfrac{dy}{dx}\), \(f'(x)\), or \(\dfrac{d}{dx}[f(x)]\).
3. Gradient Function and Gradient at a Point
A gradient function gives the slope of the curve at any general point \(x\). Substitute the \(x\)-coordinate into \(f'(x)\) to find the gradient at a specific point.
4. Derivative of Algebraic Sums and Differences
\(\dfrac{d}{dx}[f(x)\pm g(x)]=\dfrac{d}{dx}[f(x)]\pm\dfrac{d}{dx}[g(x)]\). Rewrite roots and denominators in index notation before differentiating.
5. Chain Rule for Composite Functions
If \(y=g(u)\) and \(u=h(x)\), then \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\). For \(y=u^n\), \(\dfrac{d}{dx}(u^n)=nu^{n-1}\dfrac{du}{dx}\).
6. Product Rule
For \(y=uv\), \(\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\), or \((uv)'=uv'+vu'\).
7. Quotient Rule
For \(y=\dfrac{u}{v}\), where \(v(x)\neq0\), \(\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}\).
8. Special Cases
If \(n=1\), \(y=ax\implies\dfrac{dy}{dx}=a\). If \(n=0\), \(y=a\implies\dfrac{dy}{dx}=0\).
Steps to Find the Gradient at a Point
Find the gradient function \(f'(x)\) using differentiation rules.
Substitute the value of \(x\) into \(f'(x)\).
Simplify the numerical value.
Interpret the result as the gradient of the tangent at that point.
Basic power rule
\(\dfrac{d}{dx}(ax^n)=anx^{n-1}\)Multiply by the power, then reduce the power by 1.
Chain rule for \(y=u^n\)
\(\dfrac{dy}{dx}=nu^{n-1}\dfrac{du}{dx}\)Differentiate the outer shell, then multiply by the inner derivative.
Try to Answer First
Answer in your mind, then press "Check Answer".
Differentiate \(f(x)=4x^3-\dfrac{2}{x^2}\) with respect to \(x\).
Given \(y=(3x^2-1)^4\), find \(\dfrac{dy}{dx}\).
Find the gradient of the tangent to the curve \(y=\dfrac{x}{2x+1}\) at \(x=1\).
Common Mistakes
- Differentiating products directly: writing \(\dfrac{d}{dx}(uv)=\dfrac{du}{dx}\times\dfrac{dv}{dx}\) instead of using the product rule \(u\dfrac{dv}{dx}+v\dfrac{du}{dx}\).
- Reversing the order in the quotient rule: writing \(u\dfrac{dv}{dx}-v\dfrac{du}{dx}\) instead of \(v\dfrac{du}{dx}-u\dfrac{dv}{dx}\). The subtraction must start with \(v\).
- Forgetting the inner derivative in the chain rule: differentiating \((2x+3)^4\) as \(4(2x+3)^3\) while forgetting to multiply by \(\dfrac{d}{dx}(2x+3)=2\).
- Incorrect conversion of radical and reciprocal terms: confusing \(\dfrac{1}{4x^3}\) with \((4x)^{-3}\) instead of \(\dfrac{1}{4}x^{-3}\), leading to incorrect coefficients.
- Confusing the gradient function with the gradient at a point: leaving the answer as an algebraic expression when asked for the gradient at a specific numerical point.
Concept Misunderstandings
\(\dfrac{dy}{dx}\) is a regular fraction where \(dy\) is divided by \(dx\).
While \(\dfrac{dy}{dx}\) behaves like a ratio in chain rule notation, it represents an operator meaning "the derivative of \(y\) with respect to \(x\)" or the limiting value of \(\dfrac{\delta y}{\delta x}\) as \(\delta x\to0\).
The derivative of any expression with \(x\) can be found simply by reducing the visible power by 1.
The power rule only applies directly to single terms of the form \(ax^n\). Composite powers, products, and fractions require the chain rule, product rule, and quotient rule respectively.
The quotient rule can be written as \(\dfrac{du/dx}{dv/dx}\).
The derivative of a quotient is never the quotient of the individual derivatives. The formula \(\dfrac{vu'-uv'}{v^2}\) must be applied.
Summary
- The basic power rule is \(\dfrac{d}{dx}(ax^n)=anx^{n-1}\). Constants differentiate to 0, and linear terms \(ax\) differentiate to \(a\).
- The derivative function \(f'(x)\) gives the gradient function of the curve \(y=f(x)\). Substituting an \(x\)-value yields the gradient of the tangent at that point.
- The chain rule differentiates composite functions: \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\), or \(\dfrac{d}{dx}(u^n)=nu^{n-1}\dfrac{du}{dx}\).
- The product rule states that \(\dfrac{d}{dx}(uv)=uv'+vu'\).
- The quotient rule states that \(\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{vu'-uv'}{v^2}\) where \(v\neq0\).
Short Activity
Objective: Identify the first derivative, chain rule, product rule, quotient rule and key terms in differentiation.
A. Objective Quiz
1 What is the first derivative of \(y=3x^4-5x+7\) with respect to \(x\)?
2 Given \(y=(2x-3)^5\), what is \(\dfrac{dy}{dx}\)?
B. Fill in the Blanks
3 The formula for the product rule when differentiating \(y=uv\) with respect to \(x\) is \(\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+\) ________.
4 If \(f(x)=x^2\), the numerical gradient of the curve at the point \((2,4)\) is ________.
C. Matching / Drag and Drop
Drag each differentiation rule or concept to the correct mathematical expression. On a phone, tap a term first, then tap the matching box.
