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Differential equations
A differential equation is a type of equation that involves derivatives. In other words, a differential equation represents a situation where the rate of change of a quantity is dependent on the current state of the quantity.
Differential equations are broadly classified into two types:
Ordinary Differential Equations (ODEs): These involve derivatives with respect to only one variable. They are further classified based on the order (the highest derivative in the equation) and the degree (the power of the highest derivative in the equation).
Partial Differential Equations (PDEs): These involve derivatives with respect to more than one variable.
Differential equations are typically represented symbolically. For example, an ordinary differential equation might be written as: \[f'(x) = \frac{dy}{dx}\]
This is equivalent to 'change in y divided by change in x'. Variables x and y can be substituted for any other letter.
In the case of partial derivatives, the "d"s would be instead represented by the ∂ symbol: \(f'(x) = \frac{∂y}{∂x}\).
The apostrophe (') written behind the letter symbolising a function denotes that the equation that follows is not the original equation but its derivative. Differentiating a function y 'with respect to x' (meaning x is the value on the bottom of the fraction) results in the derivative y '. If the function is represented as f (x), then its derivative can be represented as f'(x).
Let's do a quick review of how to find the gradient of a straight line graph:
Example of a gradient of a straight line graph,
However, if we look at a quadratic graph, it isn't clear how to find its gradient. This is because it changes at different points in the graph as the line curves, getting more or less steep.
One potential method we could use is to draw a tangent at a given point and find its equation. However, this would only give us the gradient at that point - what if we wanted to find a general expression for the gradient of any point on the graph?
We use differentiation to find a function for the gradient of a graph. The method is very straightforward - you need to:
Decrease the power of x by one
Multiply by the old power
Therefore, as a general rule, when differentiating xn, your result is \(nx^{n-1}\).
How do you differentiate a polynomial?
Let's say we have the following graph of \(y = x^2 + x+2\) and we want to find the gradient at the point x = 1.
To differentiate the function, we take each power of x and perform the above steps on it - reduce the power by 1, and multiply by the old power.
\(y = x^2 + x +2\)
\(x^2 \Rightarrow 2x^{2-1} = 2x\)
\(x \Rightarrow x^{1-1} = x^0 = 1\)
2 isn't a power of x, so we can't apply our usual method here.
To understand how to differentiate it, we need to look at the representation of differentiation \(\frac{dy}{dx}\). As a reminder, this means 'the change in y divided by the change in x'.
Since 2 is a constant, changes in x and y do not affect its value, and vice versa. This effectively means that for the gradient it doesn't matter what the value is - it is only important in the context of the original function. For this reason, the derivative of a constant is defined as 0.
Now that we have found the derivative of each of the terms in our function, we can create a function for the gradient at any given point:
\(y = x^2 + x+2\)
\(\frac{dy}{dx} = 2x +1\)
Therefore to find the gradient at the point where x = 1, substitute this value into our new equation:
\(m = 2(1) + 1 = 3\)
What is differentiation from first principles?
Differentiation from first principles tells us about the concept of differentiation.
Let's consider this curve which is part of a graph that we would like to differentiate. We have chosen two points along it, (x, f (x)) and (x + h, f (x + h)), and we would like to find the gradient at the point (x, f (x)):
We know to find the gradient between these points, we find the change in y divided by the change in x:
\(m = \frac{f(x+h) - f(x)}{x+h-x} = \frac{(x+h)-f(x)}{h}\}\)The closer we move those two points together, the better our estimate of the gradient at (x, f (x)) will be. As h gets closer and closer to 0, the estimate will be better and better. We can write this as the formula:
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
We know the derivative of x2 is 2x, but we can prove this by substituting it into the formula:
\(f(x) x^2\)
\(f'(x) = \lim_{h \rightarrow 0} \frac{(x+h)^2-x^2}{h} = \lim_{h \rightarrow 0} \frac{x^2+2xh+h^2-x^2}{h} = \lim_{h \rightarrow 0}\frac{h(2x+h)}{h} = \lim_{h \rightarrow 0} 2x + h\)
\
Finally, we need to consider what happens at the limit as h approaches 0: h disappears, and we are just left with our answer 2x.
What can differentiation tell us about graphs?
Differentiation can tell us a lot about the nature of graphs and their turning points. These are also known as critical points as they are points where the gradient is equal to zero. There are three possibilities when this is the case:
When the graph is quadratic, it's obvious if the critical point is a maximum or minimum, as there is only one, and all you need to do is consider the shape of the graph (using the coefficient of the x2 term). However, when there are multiple critical points, it isn't so clear.
In order to determine the nature of a critical point for cubic graphs, you need to check the gradients on either side of it.
Let's consider a local maximum:
We can see that the first part of the graph is increasing according to the direction of the graph, then after the critical point, it starts to decrease.
If we found the gradient of the increasing part of the graph, it would be positive, and the decreasing part would be negative. In summary:
\[\frac{dy}{dx} > 0 \quad \text{increasing}\]
\[\frac{dy}{dx} = 0 \quad \text{critical point}\]
\[\frac{dy}{dx} < 0 \quad \text{decreasing}\]
Let's look at determining the nature of a critical point.
\(y = x^2 + 4x +2\)
We already know that the critical point of this graph is going to be a minimum, because the x2 has a positive coefficient. However, we'll prove it using differentiation.
First, we need to differentiate the function;
\(y' = 2x + 4\)
Now we need to find the coordinates of the critical point, the x value where the derivative of the function is zero. We can do this by solving the equation \(2x + 4 = 0\), since we know the gradient is zero at that point.
\(2x + 4 = 0 \rightarrow 2x = -4 \rightarrow x = -2\)
Now we can create a simple table and sub in the values of x on either side:
| x = -3 | x = -2 | x = -1 |
| x' = 2(-3) + 4 = -2 | x' = 0 | x' = 2(-1) + 4 = 2 |
| Negative so decreasing | Turning point | Positive so increasing |
Since the gradient on the left is decreasing and the gradient on the right is increasing, we have shown that the turning point is a minimum.
If the gradient on the left would be increasing and the gradient on the right decreasing, the turning point would be a maximum.
Finally, if they are both increasing or both decreasing, it must be a stationary point.
What can the second derivative tell us about graphs?
A different possibility to determine if a critical point is a maximum, minimum, or stationary point is by using the second derivative, as the second derivative of a graph tells you its curvature.
A positive curvature means the graph curves towards the left if considered along the x-axis (minimum).
A negative curvature means that the graph curves towards the right (maximum).
If the second derivative of a function is zero at a certain point, the curvature is zero, and the graph is straight at this point (stationary point).
In our example:
\(y = x^2 + 4x+2\)
\(y' = 2x +4 \)
\(y'' = 2\)
This means that the curvature is positive anywhere on the graph and the critical point is a maximum.
Differentiation rules
Some differentiation rules which help you to find the derivatives of more complex functions are:The Product Rule
The Quotient rule
The Chain rule
The product rule
The product rule can be used to find the derivative of two functions multiplied together. The formula is;
If y = uv, then \(y' = uv' + vu'\)
Where u is the function f(x) and v is the function g(x), and f'(x), g'(x) are their derivatives u' and v'.
Differentiate the function \(y = (x^2 + 1)(x^2+x)\)
We could expand the brackets in this example and find the derivative the usual way, however often using the product rule is faster and less prone to error.
To use the product rule on this function, we need to let \(u = x^2 + 1\) and \(v = x^2 + x\)x.
Next, we need to differentiate them individually:
\(u' = 2x\)
\(v' = 2x+1\)
Finally, we substitute these values into the product formula:
\(y' = (x^2 + 1)(2x+1) + 2x(x^2+x) = 2x^3 + x^2 + 2x + 1 + 2x^3 + 2x^2 = 4x^3 + 3x^2 + 2x +1\)The quotient rule
The quotient rule can be used to find the derivative of two functions divided by each other. The formula is: \(y = \frac{u}{v}\)\(y' = \frac{vu' -uv'}{v^2}\)
Where u is the function f (x) and v is the function g (x), and f '(x), g' (x) are their derivatives u' and v'.
Differentiate the function \(y = \frac{x}{x+2}\)
We let u be the numerator, and v be the denominator, ie u = x and v = x + 2, then differentiate them individually as before to get u' = 1 and y' = 1.
Finally, we need to substitute these values into the formula:
\(y' = \frac{(x+2)(1) - (x)(1)}{(x+2)^2}\)
\(y' = \frac{2}{(x+2)^2}\)
The chain rule
The chain rule can be used to find the derivative of a function of a function. The formula is;
\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)Differentiate the function \(y = (x+2)^3\)
We let \(u = x+2\), then substitute this into the main equation such that \(y = u^3\). We then differentiate them both individually, thus finding \(\frac{dy}{du}\) and \(\frac{du}{dx}\);
\(\frac{du}{dx} = u'(x) = 1\)
\(\frac{dy}{du} = y'(u) = 3u^2\)
Finally, we multiply them together to get \(\frac{dy}{du} = 3u^2\) , and substitute u back in to get \(y' = 3(x+2)^2\).
Parametric differentiation
Sometimes we want to differentiate functions where x and y are both in terms of a third variable. In these situations, we need to use parametric differentiation.
\(y = 3t^2 + 2t -3\)
\(x = 4t + 5\)
We can use the chain rule to differentiate in terms of x and y:
\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
We could rearrange the equation involving x to be in terms of t. The above equation could also be written as the following, making it easier to differentiate:
\(\frac{dy}{dx} = \frac{dy}{dt}/\frac{dx}{dt}\)
Let's first try rearranging and multiplying our results: \(x = 4t + 5 \Rightarrow t = \frac{1}{4} (x-5) \rightarrow t' = \frac{1}{4}\)
\(y = 3t^2 + 2t - 3 \rightarrow y' = 6t + 2\)
\(\frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{4} (6t +2) = \frac{3t +1}{2}\)
Now let's try the second method to ensure we get the same answer. All we need to do is differentiate each equation individually with respect to t, and then divide \(\frac{dx}{dt}\) by \(\frac{dy}{dt}\):
\(y = 3t^2 + 2t - 3 \rightarrow y' = 6t + 2\)
\(x = 4t + 5 \rightarrow x'=4\)
\(\frac{dy}{dx} = \frac{6t+2}{4} = \frac{3t+1}{2}\)
Implicit differentiation
When differentiating, we are usually faced with explicit functions - that is, functions which generally look like \(y = x^2 + 3x +...\) . However, what if we wanted to differentiate the equation \(x^2 + y^2 = 25\)?We need to use a technique called implicit differentiation to solve this. We can approach each part of the equation separately and write:
\(\frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx} 25\)
We know how to differentiate two of the parts. The first stage to differentiating the y part is to differentiate it as normal, but leave \(\frac{dy}{dx}\);
\(2x + \frac{dy}{dx} 2y = 0\)
Now we need to rearrange the equation in terms of \(\frac{dy}{dx}\):
\(\frac{dy}{dx}2y = -2x\)
\(\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\)
Differentiation - key takeaways
Differentiation is a method of finding rates of change, i.e. gradients of functions.
The result of a differentiation calculation is called the derivative of a function.
The process of differentiation is represented by \(\frac{dy}{dx}\).
- To differentiate a polynomial:
Decrease the power of x by one
Multiply by the old power
- The derivative of a constant is defined as 0.
- Differentiation from first principles uses the formula, \(f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
- \(\frac{dy}{dx} > 0\) increasing
- \(\frac{dy}{dx} = 0\)critical point
When the derivative is equal to zero, there are three possibilities:
\(\frac{dy}{dx} < 0\) decreasing
The product rule is \(y'= uv'+vu'\)
The quotient rule is \(y' = \frac{vu'-uv'}{v^2}\)
The chain rule is \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
Parametric differentiation uses the formula \(\frac{dy}{dx} = \frac{dy}{dt}/\frac{dx}{dt}\)
Implicit differentiation involves differentiating each part of the equation separately and rearranging for \(\frac{dy}{dx}\)
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Frequently Asked Questions about Differentiation
How do you differentiate a fraction?
To differentiate a fraction, you need to use the quotient rule;
y'=(vu'-uv')/v^2
What is differentiation?
Differentiation is the process of finding a function for the gradient of a given function.
How do you differentiate a function of a function?
To differentiate a function of a function, you need to use the chain rule; dy/dx=dy/du ⋅ du/dx
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