Aproximações lineares e diferenciações

In the previous section, we learned about linear approximations for a function $f(x)$ around $x=a$: $$f(x)\approx f(a)+f'(a)(x-a).$$ This idea is sometimes stated in the perspective and terminology of differentials:

For the function $y=f(x)$ above, differentiable at $x=a$, the differentials $dx$ and $dy$ denote changes in $x$ and $y$ of the tangent line to the graph of $f(x)$ at $x=a$. Since the tangent line has slope $f'(a)$, we then have $$f'(a)=\frac{\text{rise}}{\text{run}}=\frac{dx}{dy},$$ where the expression on the right is an actual fraction, and not the notation for differentiation. Now, we can find a formula for the dependent differential $dy$: $$dy=f'(a)\, dx.$$ Of course, $dx$ is the independent variable in this equation.

Now, let's return to our earlier notion of linear approximation: $$f(x)\approx f(a)+f'(a)(x-a).$$ Can you find a way to rewrite this equation in terms of the differentials $dx$ and $dy$?

Let's say $dx=x-a$. Then this implies $x=a+dx$, and so $$f(a+dx)=f(x)\approx f(a)+f'(a)(x-a)\\=f(a)+f'(a)\, dx=f(a)+dy.$$

Great! Now, not only have we learned what differentials are, we also related it to our notions of linear approximation! This calls for some practice problems:

Since $f'(x)=2x$, we have $$dy=f'(3)\, dx=2\cdot 3\, dx=6\, dx.$$

That was a bit simple, so let's try a harder problem.

Let's let $f(x)=\sqrt{x}$. In our case, since $4.02=4+0.02$, we'll set $a=4$ and $dx=0.02$. Then, since $f'(x)=\frac{1}{2\sqrt{x}}$, we have $$dy=f'(a)\, dx=\frac{dx}{2\sqrt{a}}=\frac{0.02}{2\sqrt{4}}=0.005.$$ Therefore, our linear approximation is $$f(a+dx)\approx f(a)+dy=\sqrt{4}+0.005=2.005.$$

One last thing: although the terms and notations seem new, there is nothing conceptually different between this perspective and the previous approach for linear approximation. The notion of differentials just gives us another way to describe the process.

(Video) Using Differentials

by PatrickJMT

Here's a video by patrickJMT showing you how to do a linear approximation using differentials.

The differential formula $dy$ is: $$\begin{align} \text{if, }\ y&=f(x)\\ \text{then, }\ dy&=f'(x)dx \end{align}$$

Th key idea to remember when facing these problems is that $$dy\approx\Delta y$$ where $\Delta y$ is the true change in value and $dy$ serves as an approximation. So we can approximate the true change by using the differential formula above.

The problem that he solve is using differentials to approximate the value $(18)^{\frac14}$ or $\sqrt[4]{18}$.

To begin, we're trying to find the value of $18^{\frac14}$. In a more general case, we're trying to find the the value of some number to the $\frac14$ power. So the function we would use is: $$f(x)=x^{\frac14}$$

Conceptually, we have our graph of $x^{\frac14}$ and we we have our point $(18,18^{\frac14})$.

What we want to do is pick a number close to 18 that we can easily take the fourth root of. In this case, $16$ is that number. If we take the fourth root of $16$ we get $2$; $\sqrt[4]{16}=2$.

This change in height, going from $2$ to $18^{\frac14}$, is what we call $\Delta y$ or the true change in $y$.

So we can write that $18^{\frac14}$ is equal to the prior $y$ value plus whatever $\Delta y$ is. $$18^{\frac14}=y+\Delta y$$ But we don't know exactly what $\Delta y$ is so we can approximate the value using $dy$. $$18^{\frac14}\approx y+dy$$

We need to take the differential of the function $f(x)=x^{\frac14}$ now. $$dy=\frac14x^{-\frac34}dx$$ This gives us a formula for our $dy$ value. So, we need to find an $x$ and $dx$ to plug in.

Our $x$ will be the "easy to evaluate number" that we used, $x=16$, and the $dx$ will be the change between the two $x$'s, $dx=18-16=2$.

Plug these values into differential of the function. $$\begin{align} dy&=\frac{1}{4\sqrt[4]{x^3}}dx\\ &=\frac{1}{4\sqrt[4]{(16)^3}}(2)\\ &=\frac{1}{2(2^3)}\\ &=\frac{1}{2(8)}\\ &=\frac{1}{16} \end{align}$$

Now that we have a value for $dy$ we can plug that into the approximation equation. $$\begin{align} 18^{\frac14}&\approx2+\frac{1}{16}\\ &\approx\frac{32}{16}+\frac{1}{16}\\ &\boxed{\approx\frac{33}{16}} \end{align}$$

Let's see how accurate this was. $$\frac{33}{16}=2.0625\\ \sqrt[4]{18}=2.0598$$ That's pretty accurate!