Differentiation From First Principles
Differentiation From First Principles – Introduction
Differentiation from first principles, also known as the limit definition of derivative, is a foundational concept in calculus. This concept is applicable to A Level Maths students and is used to compute the derivative of a function. It involves finding the rate of change of a function at a specific point by analysing its behaviour as the change in the input approaches zero.
To understand differentiation from first principles, it is helpful to briefly explore the history of calculus and its development. The concept of differentiation was first introduced by Sir Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century. They independently discovered the fundamental principles of calculus, including differentiation and integration, which revolutionised mathematics and other scientific disciplines.
Now, let’s dive into the concept of differentiation from first principles. Suppose we have a function f(x), and we want to find its derivative at a specific point x=a. The derivative of a function represents its instantaneous rate of change at a given point. To compute the derivative using first principles, we follow these steps:
Choose a small value h (approaching zero) to determine the change in the input variable (Δx).
Calculate the difference quotient (Δy/Δx) by evaluating the function at x=a+h and x=a.
Take the limit of the difference quotient as h approaches zero to obtain the derivative.
Mathematically, the derivative of a function f(x) at x=a can be expressed as:
f'(a) = lim(h→0) {f(a+h) – f(a)}/h
Applying this definition, we can find the derivative of various functions by substituting the appropriate values into the formula. This approach provides an accurate and precise calculation of the derivative, especially for complex functions where other methods may not be readily applicable.
Understanding differentiation from first principles is crucial as it forms the foundation for more advanced concepts in calculus. It helps students grasp the concept of instantaneous rate of change, which has numerous applications in physics, engineering, economics, and other fields.
In summary, differentiation from first principles is a fundamental concept in calculus that A Level Maths students need to comprehend. It involves using the limit definition of derivative to find the rate of change of a function at a specific point. This concept was initially developed by Newton and Leibniz, and it revolutionised the field of mathematics. By understanding and applying differentiation from first principles, students can effectively calculate derivatives and explore the intricacies of calculus.
Differentiation From Principles - 2 Examples
Example 1
When it comes to differentiation from first principles, the main area of maths that is being used is expansion and collecting of like terms. Now depending on the power within a question you may need to use Binomial Expansion. You also need to be careful with any signs otherwise you could end up with additional terms that you do not need.
In all cases, the process is the same, the only thing that will change is the amount of algebra and its intensity.
\begin{aligned} & \frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{2(x+h)^2-5(x+h)+3-\left[2 x^2-5 x+3\right]}{h} \\ & =\lim _{h \rightarrow 0} \frac{2\left(x^2+2 x h+h^2\right)-5 x-5 h+3-2 x^2+5 x-3}{h} \\ & =\lim _{h \rightarrow 0} \frac{2 x^2+4 x h+2 h^2-5 x-5 h+3-2 x^2+5 x-35}{h} \\ & =\lim _{h \rightarrow 0} \frac{4 x k+2 h^2-5 k}{k} \\ & =\lim _{h \rightarrow 0} 4 x+2 h-5 \end{aligned}\text { As } h \rightarrow 0 \quad 2 h \rightarrow 0 \quad \therefore \frac{d y}{d x}=4 x-5
Even though you have an unknown constant you still need to perform the differentiation as you would normally and this is shown below. You can then substitute in the additional information that you have in order to find the value of the constant as shown below:
\frac{d y}{d x}=-a x^{-2}+3 x^{\frac{1}{2}}When x = 4:
\begin{aligned} & =-a(4)^{-2}+3(4)^{\frac{1}{2}} \\ & =-\frac{9}{4^2}+3 \sqrt{4} \\ & =-\frac{9}{16}+(3 \times 2) \\ & =-\frac{9}{16}+6 \end{aligned}\begin{aligned} \frac{d y}{d x}=7 \Rightarrow-\frac{a}{16}+6 & =7 \\ -a+96 & =112 \\ a & =112-96=-16 \end{aligned}Example 2
The good thing about doing differentiation from first principles is that you know what the answer using the standard result.
\begin{aligned} & \frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{(x+h)^2-3(x+h)+4-\left[x^2-3 x+4\right]}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left(x^2+2 x h+h^2-3 x-3 h+4-x^2+3 x-4\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{2 x k+h^k-3 k}{k} \\ & =\lim _{h \rightarrow 0} 2 x+h-3 \\ & \text { As } h \rightarrow 0, \quad h \rightarrow 0 \quad \therefore \frac{d y}{d x}=2 x-3 \end{aligned}
For the above question you first need to rewrite the question in index form so you can perform the differentiation.
Differentiation From Principles - 2 More Examples
Example 3
Example 4
To perform this question you first need to rewrite the expression in a form that can be differentiated and this is shown below:
\begin{aligned} y & =3 x^{-2}+x^{\frac{5}{2}} \\ \frac{d y}{d x} & =(-2)(3) x^{-2-1}+\frac{5}{2} x^{\frac{5}{2}-1} \\ & =-6 x^{-3}+\frac{5}{2} x^{\frac{3}{2}} \\ & =-\frac{6}{x^3}+\frac{5}{2} x^{\frac{3}{2}} \end{aligned}
Differentiation From Principles – Final 2 Examples
Example 5
Example 6
When x = 4
\begin{aligned} & =\frac{1}{2}(4)^{-\frac{1}{2}}-k\left(4^{-2}\right) \\ & =\frac{1}{2 \sqrt{4}}-\frac{k}{4^2} \\ & =\frac{1}{4}-\frac{k}{16} \end{aligned}When \frac{d y}{d x}=2
\begin{aligned} \frac{1}{4}-\frac{k}{16} & =2 \\ 4-k & =32 \\ k & =4-32=-28 \end{aligned}All the questions did not involve a power higher than 2 but as mentioned earlier if the power is higher than 2, then you are more than likely to need to use Pascal Triangles in order to perform the expansion.
Remember that the hardest part with differentiation from first principles will be the expansion and simplification of terms.