Continuous Change (Calculus)
Key Points
The derivative of \(y\) with respect to \(x\) is given by
\begin{align}\frac{dy}{dx}\end{align}
and corresponds to the “slope function” of tangents to a curve on the \(x\)-\(y\) plane.
The derivative of \(f\) with respect to \(x\) can be found using:
\begin{align}\frac{df}{dx}=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{g}\end{align}
and is known as differentiating from first principles.
The derivative of any function of the form \(y(x) =A\), where \(A\) is not a function of \(x\), is given by
\[\frac{dy}{dx}= 0\]
The derivative of any function of the form \(y(x) =x^n\) is given by
\[\frac{dy}{dx}= nx^{n-1}\]
The derivative of a function of the form
\[y(x) = au(x)\]
is given by
\[\frac{dy}{dx} = a\frac{du}{dx}\]
where \(a\) is a constant and \(u(x)\) is a function with a known derivative.
The derivative of a function of the form
\[y(x) = u(x) \pm v(x)\]
is given by
\[\frac{dy}{dx} = \frac{du}{dx} \pm \frac{dv}{dx}\]
where \(u(x)\) and \(v(x)\) are functions with known derivatives.
The derivative of a function of the form
\[y(x) = u(x)v(x)\]
is determined by using the product rule
\[\frac{dy}{dx} = u\frac{dv}{dx}+v\frac{du}{dx}\]
where \(u(x)\) and \(v(x)\) are functions with known derivatives.
The derivative of a function of the form
\[y(x) = \frac{u(x)}{v(x)}\]
is determined by using the product rule
\[\frac{dy}{dx} = \frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}\]
where \(u(x)\) and \(v(x)\) are functions with known derivatives.
To differentiate a function that is either a product of functions with known derivatives or a ratio of functions with known derivatives, we perform the following algorithm:
- Step 1: Set \(u\) and \(v\) as each of the known functions.
- Step 2: Calculate the derivative of \(u\) and the derivative of \(v\).
- Step 3: Calculate the derivative of the original function by inserting the above into either the product rule or the quotient rule.
The derivative of a function of the form \(y(u(x))\) is determined by using the chain rule
\[\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\]
where \(u(x)\) is a function with a known derivative.
To differentiate a function with respect to \(x\) that contains a nested function with a known derivative, we perform the following algorithm:
- Step 1: Let \(u\) be the nested function. Write the original function in terms of \(u\).
- Step 2: Calculate the derivative of the original function with respect to \(u\) and the derivative of \(u\) with respect to \(x\).
- Step 3: Calculate the derivative of the original function with respect to \(x\) using the chain rule.
- Step 4: Reinsert what was defined in step 1 as \(u\).
The “derivative of a derivative” is know as the second derivative as is given by
\[\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right)\]
The integral of any function of the form \(y(x) =x^n\) is given by
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} +C\]
Let \(I(x)\) be the indefinite integral of the function \(y(x)\), i.e.
\[I(x) = \int y(x)\,dx\]
The definite integral of this function between \(x=a\) and \(x=b\) is then given by
\[\int_a^b y(x)\,dx=I(b)-I(a)\]
and corresponds to the area between \(I(x)\), the \(x\)-axis and the lines \(x=a\) and \(x=b\).
The integral of a function of the form
\begin{align}y(x) = au(x)\end{align}
is given by
\begin{align}\int y(x)\, dx &=\int au(x)\,dx \\&= a\int u(x) \, dx \end{align}
where \(a\) is a constant and \(u(x)\) is a function with a known integral.
The integral of a function of the form
\[y(x) = u(x) \pm v(x)\]
is given by
\begin{align}\int y(x) \, dx &= \int (u(x)\pm v(x))\, dx\\&= \int u(x) \, dx \\&\pm \int v(x) \, dx\end{align}
where \(u(x)\) and \(v(x)\) are functions with known integrals.
The average valune of a function \(y(x)\) between \(x=a\) and \(x=b\) is given by:
\begin{align}\frac{1}{b-a} \int_a^by(x)\,dx\end{align}
Causes of Confusion
You will often see a variable \(y\) that depends on \(x\) instead written as \(y(x)\), e.g. \(y(x) = x^2\).
\(y(x)\) is referred to as the dependent variable as its value depends on what the value of \(x\) is. \(x\) is instead referred to as the independent variable.
Note that \(y(x)\) does not mean \(y\) multiplied by \(x\). Instead, the \(x\) in brackets is there simply to remind us that the value of \(y\) depends on what the value of \(x\) is. We also therefore often say that \(y\) is a function of \(x\).
Note that, on the page containing known integrals in the Formulae and Tables booklet, it states “Constant of integration omitted.”
This is purely for neatness purposes so that the integrals do not take up more space on the page.
It does not, however, mean that you can omit the constant of integration when answering a question. The constant of integration must be included when calculating indefinite integrals.
Exam Tips
When finding derivatives of functions, you may have to use both e.g. the product rule and the chain rule for different parts of the function. Therefore, what is meant by \(u\) when using the product rule on one part of the function is not what is meant by \(u\) when using the chain rule on another part of the function.
In order to avoid this confusion, it is best to instead use different letters when applying different rules within the same question.
Additional Resources
Derivative plotter – graphs of both a function and its derivative
The paradox of the derivative – an informative video on derivatives as applied to the motion of a car
Definite Integral Calculator – visualise the relationship between integrals and areas for any function
