Lecture 1. Differential calculus of functions of a single variable
We start studying a section called differential calculus of functions of a single variable. The key concept of this section is the concept of a derived function. These concepts were introduced by the famous mathematicians Isaac Newton, Gottfried Leibniz who lived in the XVII century.
Let’s start with the concept of a derived function at a point. The function f is defined at the point x0 and in some of its neighborhood. We change the value of the function x0 to some non-zero value Δ(Delta) x, which is commonly called the increment of the argument, and we get a new point on the abscissa axis, the point x0+Δx. What happens to the function f? Its new value on the y-axis at the point x0 + Δx, the value by which the value of the function has changed, is called the increment of the function f at the point x0, denoted by Δf(x0). Look how this value is calculated (see the video).
If there is a finite limit to the ratio of the increment of the function at x0 to the increment of the argument when it tends to zero, the value of this limit is called the derivative of the function f at x0. The function f is called differentiable at x0. f’(x0) is a common notation for the derivative. In addition, when solving problems, we will use other notations (see the video). All these are the notations of the derivative of the function f at the point x0.
Keep in mind that sometimes a function is not differentiable. These are cases when the specified limit either does not exist or is equal to infinity. What is a necessary condition for differentiability? Let’s formulate a theorem. If the function is differentiable, it is continuous at this point. Continuity is a necessary condition for differentiability.
Let’s prove it. It’s not complicated. The function is differentiable at a point, so at this point x0 there is a finite derivative, which is defined as the limit. To establish continuity, we need to use the definition. Look, a function is called continuous if the limit of the increment of functions when tending to zero is zero.
Let’s calculate the limit of the function increment. What will we do for this? We divide the increment of functions by Δx and multiply it by Δx and apply the product limit theorem. The limit of the first multiplier is f’(x0), the limit of the increment is zero, we get zero. Based on the definition of continuity in the language of increments, we conclude that the function is continuous at the point x0. The theorem is proved.
Note that the inverse theorem is not true, that is, the classes of differentiable functions and continuous functions do not coincide. There are continuous functions that are not differentiable.
An example of such a function is a very simple function f(x) = |x|. Let’s consider the point zero as x0. It is fairly obvious that the function is continuous at point 0. Let’s assume that it is not differentiable at point 0. Let’s try to calculate the derivative. If the derivative exists at point 0, it is calculated as the limit of the ratio of the function increment at point 0 to the increment of the argument Δx when the latter tends to zero.
Let’s try to calculate this limit. First, let’s find out what Δf is from zero. By the formula (see the video), we calculate the increment and see that it is |Δx|. It is obvious that it will open differently depending on the sign of Δx. The former case. Let Δx>0, that is, the increment of the function is Δx. By calculating the limit of the ratio for Δx>0, we get that the limit is equal to one.
The second case. We calculate the same limit for Δx<0. What do we get? The limit of the ratio is equal to -1. So, the limit of the ratio of the increment of the function at point 0 to the increment of the argument Δx, when Δx tends to zero, depends on whether Delta Δx>0 or Δx<0, that is, one-sided limits are not equal. The conclusion is that the specified limit does not exist, which means that the derivative at point 0 does not exist. The function is continuous at point 0, but differentiability at point 0 is broken.
How do we determine from a graph that the function is not differentiable? Graphs of differentiable functions are called smooth. At point 0, the smoothness of this curve is broken, and the function is not differentiable, despite the fact that it is continuous.
Now let’s try to calculate the derivatives of a function using the definition. The task is to find the derivative of a function at an arbitrary point x for a constant function. f (x)=c at any point x of the numeric line. Let’s make up the definition. x is an arbitrary point of the numeric line. We find the increment of the function – the numerator of the fraction. We give the argument at any point x an increment of Δx. We get a new point x0 + Δx.
What do we see? Since the value of the function at any point is c, the increment of the function is always zero. What do we get as a result? We have to calculate the limit of the fraction 0 / Δx. This function is a constant everywhere except where Δx is zero. The limit of the constant is zero. We conclude that the derivative of the constant at any point is zero.
Let’s take another example. The task is to find the derivative of the function sin(x) at any point x of the numeric line. We write down the definition again and give the argument x a non-zero transformation Δx. We make an increment of the function. Next, for convenience, we use the formula for the difference of sines and convert Δf(x) to a new form. Now let’s calculate the limit. We got the ratio under the limit sign (see the video). We see here the uncertainty 0 / 0, trigonometry, so we use the first remarkable limit. Note that this expression, this specified fraction (see the video) on the first remarkable limit tends to one. Further calculating the limit, Δx tends to zero, we get cos(x). we conclude that the derivative sin(x) at any point of the numeric line is equal to cos(x).
We can continue calculating the derivatives of any functions by definition, and as a result we will get the facts that are easily combined into a table of derivatives. This is what you need to know to calculate derivatives. For reference, you can see the latest formulas in this table that relate to hyperbolic functions. So, they are called hyperbolic sine, cosine, tangent and cotangent. The tasks to calculate the derivatives in the textbooks are often associated with these functions. The reference material on the right tells you about these functions. It’s for you to know what it stands for.
Calculating the derivative as a limit can sometimes be a very time-consuming task and probably not necessary, so we turn to the differentiation rules that simplify the task of calculating derivatives. And the first theorem is on arithmetic operations on derivatives. If the functions U and V are differentiable at x0, at this point, the sum, difference, product, and quotient of the functions are differentiable, the quotient under certain restrictions. Formulas for calculating derivatives at the point x0 are generally well known and are calculated as limits. It is not complicated. We won’t do it. We remember these rules, and we will use them when solving tasks to find the derivatives.
The consequences which also can be convenient are as follows. First: a constant multiplier can be taken out of the sign of the derivative. Second: if we had a formula for the derivative of the product of two multipliers, it can easily be generalized to the product of any finite number of functions. Here, the method of mathematical induction is used to conduct a proof. Look how this formula works in this case (see the video).
In addition, functions can be connected by a composition operation, so a theorem on how to calculate the derivative of a composition of two functions is needed. If the function u is differentiated at the point x0, and the function f(u) is differentiable at the corresponding point u0, the composition of these functions the function F is differentiable at the point x0 and the derivative is calculated by the formula (see the video).
Note that if at least one of the functions u or f is a constant, the right part of the function F becomes a constant, the derivative of F will be zero and the right part will also be zero, that is, if at least one function is a constant, the formula is obviously true.
To prove this, we assume that these are not constants. Give the argument x an increment of Δx. What happens to the function u? It will get an increment of Δu and the derivative of the function u at the point x0 is calculated as the limit of the ratio of Δu to Δx, when Δx tends to zero. But Δu, in turn, is an increment of the argument for the function f. Incrementing the argument changes the value of the function f by the value Δy.
Therefore, the derivative of the function f at the point u0 is calculated as the limit of the ratio Δy to Δu, when Δu tends to zero. At the same time, Δy is an increment of the entire function F of a complex function, the composition of functions u and f. By definition, the derivative of the function F is calculated as the limit of the ratio of its increment to the increment of the argument. What else can we do? We divide by Δu, multiply by Δu, and applying the theorem on the limit of the product of functions, we get the desired conclusion about the derivative of the composition of two functions.
In practice, in order not to introduce unnecessary notation, this theorem can be written briefly as its conclusion (see the video). If u is some expression containing the variable x, the derivative is calculated as f' for the same value of the argument u and then we multiply by the derivative of this argument u. How does this formula work? Let’s look at some examples from the table of derivatives. In general, the entire table of derivatives can be rewritten again. Let’s say we have three formulas in the table of derivatives. How is the differentiation formula transformed if there is an expression instead of x? Look, we write the same formula that was previously written with x, but then the multiplier u’ always appears. So, the entire table of derivatives can be rewritten as follows (see the video).