Approximate calculations using differentials
Literature: Collection of Problems in Mathematics. Part 1. Edited by A.V. Efimov, B.P. Demidovich.
The expression for the differential is given by $$dy(x_0, dx)=f'(x_0)dx,$$ where $dx=\Delta x.$
If $f'(x_0)\neq 0,$ then as $\Delta x\rightarrow 0$, the increment of the function and its differential $dy$ at a fixed point are equivalent infinitesimals, allowing us to write the approximate equality:
$\Delta y\approx dy$ when $|\Delta x|\ll1.$
Examples.
5.298. Compute approximately:
a) $\arcsin 0.05;$
Solution.
We will use the formula $\Delta y\approx dy=y'(x_0)dx.$
$y(x)=\arcsin x$
$x_0=0,$ $\Delta x=dx=0.05,$
$\Delta y=y(x)-y(x_0)\Rightarrow \arcsin 0.05\approx y'(0)\cdot 0.05;$
$y(x)=\arcsin x\Rightarrow y'(x)=\frac{1}{\sqrt{1-x^2}}.$ Thus, $y'(x_0)=1.$
Substituting all the obtained values into the formula $\Delta y\approx dy=y'(x_0)dx,$ we get $\arcsin 0.05\approx 1\cdot 0.05=0.05;$
Answer: 0.05.
в) $\ln 1,2.$
Solution:
We will use the formula $\Delta y\approx dy=y'(x_0)dx.$
$y(x)=\ln x$
$x_0=1,$ $\Delta x=dx=0.2,$
$\Delta y=y(x)-y(x_0)\Rightarrow \ln 1.2\approx y'(1)\cdot 0.2;$
$y(x)=\ln x\Rightarrow y'(x)=\frac{1}{x}.$ Thus, $y'(x_0)=1.$
Substituting all the obtained values into the formula $\Delta y\approx dy=y'(x_0)dx,$ we get $\ln 1.2\approx 1\cdot 0.2=0.2;$
Answer: 0.2.
Homework.
5.298. Compute approximately:
b) $\arctan 1.04;$
Answer: 0.805.
5.299. Justify the approximate formula $$\sqrt[3]{x+\Delta x}\approx\sqrt[3]{x}+\frac{\Delta x}{3\sqrt[3]{x^2}}$$ and compute $\sqrt[3]{25}$ using this formula.
Answer: 2,93.
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