Approximate calculations using differentials

The expression for the differential is given by $$dy(x_0,dx)=f^{\prime }(x_0)dx,$$ where $dx=\mathrm{\Delta }x.$

If $f^{\prime }(x_0)\ne 0,$ then as $\mathrm{\Delta }x\to 0$, the increment of the function and its differential $dy$ at a fixed point are equivalent infinitesimals, allowing us to write the approximate equality:

$\mathrm{\Delta }y\approx dy$ when $|\mathrm{\Delta }x|\ll 1.$

Examples.

1. Compute approximately:

a) $\arcsin 0.05;$

Solution.

We will use the formula $\mathrm{\Delta }y\approx dy=y^{\prime }(x_0)dx.$

$y(x)=\arcsin x$

$x_0=0,$ $\mathrm{\Delta }x=dx=0.05,$

$\mathrm{\Delta }y=y(x)-y(x_0)\Rightarrow \arcsin 0.05\approx y^{\prime }(0)\cdot 0.05;$

$y(x)=\arcsin x\Rightarrow y^{\prime }(x)=\frac{1}{\sqrt{1-x^2}}.$ Thus, $y^{\prime }(x_0)=1.$

Substituting all the obtained values into the formula $\mathrm{\Delta }y\approx dy=y^{\prime }(x_0)dx,$ 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 $\mathrm{\Delta }y\approx dy=y^{\prime }(x_0)dx.$

$y(x)=\ln x$

$x_0=1,$ $\mathrm{\Delta }x=dx=0.2,$

$\mathrm{\Delta }y=y(x)-y(x_0)\Rightarrow \ln 1.2\approx y^{\prime }(1)\cdot 0.2;$

$y(x)=\ln x\Rightarrow y^{\prime }(x)=\frac{1}{x}.$ Thus, $y^{\prime }(x_0)=1.$

Substituting all the obtained values into the formula $\mathrm{\Delta }y\approx dy=y^{\prime }(x_0)dx,$ we get $\ln 1.2\approx 1\cdot 0.2=0.2;$

Answer: 0.2.

Homework.

2. Compute approximately:

b) $\arctan 1.04;$

Answer: 0.805.

3. Justify the approximate formula $$\sqrt[3]{x+\mathrm{\Delta }x}\approx \sqrt[3]{x}+\frac{\mathrm{\Delta }x}{3\sqrt[3]{x^2}}$$ and compute $\sqrt[3]{25}$ using this formula.

Answer: 2,93.

Tags: calculus, differentials, mathematical analysis