Practical lesson 1. Logarithmic differentiation. Differentiating a function defined parametrically

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Logarithmic differentiation. Differentiating a function defined parametrically

At this lesson, we continue considering the topic of differentiation of functions. It turned out that we haven’t covered the entire class of functions yet. There’s one more left.

So, this is a function U in power V where U and V are in turn functions of the variable x, such a function is called power-exponential. We can’t use the table of derivatives because we have a formula for differentiating a power function, we have a formula for differentiating an exponential function, and these formulas are not suitable for a power-exponential function. The method of logarithmic differentiation is helpful in this case. What do we need for this method? Of course, the rules of differentiation and a little knowledge of logarithms.

First, we consider the formula for the derivative of the logarithm y, if y is a function. Applying the formulas for differentiating a complex function, we see this: 1/y multiplied by y’. Let’s express y’ from this equality, we see y’ is y multiplied by the derivative of the logarithm y.

Now I want to draw your attention to this formula, and it can even be added to the list of differentiation rules, and used in these special cases. Let’s see how this new formula works. A new rule that we call logarithmic differentiation.

So, we consider the function y is Uv, we refer to the selected formula. What we write down, the derivatives of this function y’ is Uv multiplied, then the logarithm of U to the power of V and all this under the sign of the derivative. This is where we need the logarithm property. We take the exponent of V as a multiplier before the logarithm. And what we see. The problems disappeared because, at this stage, all we need to do is to find the derivative of the product of V and the logarithm of U. There are no further problems.

Now let’s use this method for a specific example. So, we see the formula. And the power-exponential function is x to the power of x. Using this formula, we write down: the multiplier y is a given function x to the power of x and then we multiply by the derivative of the logarithm of this function. We take out x as a multiplier before the logarithm and that’s it, then completely simple rules for differentiating the function are applied. So, we have derived products, we come to table functions, well, the next step is just to transform a little bit and get the answer. It is quite simple.

Let’s consider one more example. Again, this is a power-exponential function at the base of the variable x, and the exponent is also a variable x. Do not make mistakes, do not apply the formula x to the power of alpha or a to the power of x. You will make mistakes. Those formulas cannot be applied. We apply the method of logarithmic differentiation. The formula. So, we write down the first multiplier, this is y, then the logarithm of y’. The multiplier is the logarithm of x appears in front of the logarithm. And we apply the theorem on arithmetic operations. Here is the derivative of the product with primes in the first stage.

You have written it down. If you are good at differentiating, in fact, the desired entry may appear immediately, and you can even write down the answer immediately if you can perform some transformations mentally. What I can draw your attention to: logarithmic differentiation can be used not only for power-exponential functions, but also for functions that have many cofactors, especially if they are powers or exponential functions, but most often of course we use this method for logarithmic differentiation.

And we will consider the second type of problems: differentiation of a function defined parametrically. We find first-and second-order derivatives of a function given parametrically. We apply the formula, calculate the derivatives of x and y, and get the answer. For the second-order derivative, we apply the same formula. We get the following.

Let’s consider an example of a parametric function that we haven’t considered before. It is a very interesting astroid curve, how does it appear? Imagine that we have a circle of radius r. And another circle is rolling on the inner part of this circle, the radius of which is 4 times smaller. On this small circle, a point is fixed, and it leaves a trace, so the trajectory of this point is called an astroid. And you see the name is a little bit astronomical, like a star. Now, let’s see how this asteroid appears. Well, let’s look at the solution to this problem.

We find the derivative of this function given parametrically. So, we apply the formula, the derivative is calculated as y’ divided by x’, where x and y are functions of the variable t. Afterwards, we substitute data from the function system and calculate the derivative of the complex function. It is not difficult to notice that the fraction is reduced by sin, cos, and 3r, and we have the cosine t in the numerator and the denominator of the sin t and the minus sign, so minus the catangent t. That’s it, the problem is solved. 

Last modified: Понедельник, 2 декабря 2024, 2:23