Calculation of higher-order differentials.

Literature: Collection of Problems in Mathematics. Part 1. Edited by A.V. Efimov, B.P. Demidovich.

Let's consider the differential $dy(x, dx)=f'(x) dx$ as a function of $x$ at a fixed $dx$. Assuming that the function $y=f(x)$ is twice differentiable at the point $x$, let's find the differential of $dy(x, dx):$

$$d(dy(x, dx))=f''(x)dxdx$$

The value of the obtained expression is called the second differential or the second-order differential of the function $y=f(x)$ and is denoted by the symbol $d^2y(x, dx)$.

Thus, $$d^2y=f''(x)dx^2$$

Similarly,

$$d^3y=d(d^2y)=f'''(x)dx^3$$

$$.................$$

$$d^ny=d(d^{n-1}y)=f^{(n)}(x)dx^n$$

Examples:

Find the second-order differentials of the specified functions:

5.303. $y=a\sin(bx+c).$

Solution.

$y'(x)=(a\sin(bx+c))'=a\cos(bx+c)(bx+c)'=ab\cos(bx+c);$

$y''(x)=(ab\cos(bx+c))'=-ab\sin(bx+c)(bx+c)'=-ab^2\sin(bx+c).$

To compute the second-order differential, we use the formula

$$d^2y=f''(x)dx^2.$$

Thus, we obtain

$$d^2y=-ab^2\sin(bx+c)dx^2.$$

Answer: $d^2y=-ab^2\sin(bx+c)dx^2.$

5.305. $y=\frac{\sin x}{x}.$

Solution.

$$y'=\left(\frac{\sin x}{x}\right)'=\frac{(\sin x)' x-\sin x x'}{x^2}=\frac{x\cos x -\sin x}{x^2}.$$

$$y''=\left(\frac{x\cos x-\sin x}{x^2}\right)'=\frac{(x\cos x-\sin x)' x^2-(x\cos x-\sin x) (x^2)'}{x^4}=$$

$$=\frac{(x'\cos x+x(\cos x)'-(\sin x)') x^2-(x\cos x-\sin x) 2x}{x^4}=$$ $$=\frac{(\cos x-x\sin x-\cos x)x^2-2x^2\cos x+2x\sin x}{x^4}=$$ $$=\frac{-x^3\sin x-2x^2\cos x+2x\sin x}{x^4}=\frac{-x^2\sin x-2x\cos x+2\sin x}{x^3}.$$

To compute the second-order differential, the formula is known as

$$d^2y=f''(x)dx^2.$$

Thus, we obtain

$$d^2y=\frac{-2x\cos x+(2-x^2)\sin x}{x^3}dx^2.$$

Answer: $d^2y=\frac{-2x\cos x+(2-x^2)\sin x}{x^3}dx^2.$

5.307. $y=\frac{1}{x^2-3x+2}$

Solution.

Let's find $y'':$

$$y'=\left(\frac{1}{x^2-3x+2}\right)'=\left((x^2-3x+2)^{-1}\right)'=$$ $$=-(x^2-3x+2)^{-2}(x^2-3x+2)'=-(x^2-3x+2)^{-2}(2x-3).$$

$$y''=(-(x^2-3x+2)^{-2}(2x-3))'=$$ $$=-((x^2-3x+2)^{-2})'(2x-3)-(x^2-3x+2)^{-2}(2x-3)'=$$

$$=-(-2)(x^2-3x+2)^{-3}(x^2-3x+2)'(2x-3)-(x^2-3x+2)^{-2}2=$$ $$=2(x^2-3x+2)^{-3}(2x-3)^2-2(x^2-3x+2)^{-2}=$$ $$=2\frac{(2x-3)^2}{(x^2-3x+2)^3}-2\frac{1}{(x^2-3x+2)^2}=$$ $$=2\frac{4x^2-12x+9-x^2+3x-2}{(x^2-3x+2)^3}=2\frac{3x^2-9x+7}{(x^2-3x+2)^3}.$$

We find the second-order differential using the formula

$$d^2y=f''(x)dx^2.$$

Thus, we obtain

$$d^2y=2\frac{3x^2-9x+7}{(x^2-3x+2)^3}dx^2.$$

Answer: $d^2y=2\frac{3x^2-9x+7}{(x^2-3x+2)^3}dx^2.$

Find the second-order differentials of the following functions implicitly defined by $y=y(x):$

5.312. $xy+y^2=1.$

Solution.

Let's find $dy:$

$$d(xy+y^2)=d(1)\Rightarrow$$

$$ydx+xdy+d(y^2)=0\Rightarrow$$

$$ydx+xdy+2ydy=0\Rightarrow dy=-\frac{y}{x+2y}dx.$$

Next, let's find $d^2y:$

$$d(dy)=d\left(-\frac{y}{x+2y}\right)dx=-\frac{dy(x+2y)-y(d(x+2y))}{(x+2y)^2}dx=$$

$$=-\frac{(x+2y)dy-ydx-2ydy}{(x+2y)^2}dx=$$ $$=-\frac{(x+2y)\frac{-y}{(x+2y)}dx-ydx-2y\frac{-y}{x+2y}dx}{(x+2y)^2}dx=\frac{2y-\frac{2y^2}{x+2y}}{(x+2y)^2}dx^2=$$ $$=\frac{2y(x+2y)-2y^2}{(x+2y)^3}dx^2=\frac{2xy+2y^2}{(x+2y)^3}dx^2=2\frac{xy+y^2}{(x+2y)^3}dx^2.$$

Substituting the given data from the condition $xy+y^2=1,$ we finally obtain

$$d^2y=\frac{2}{(x+2y)^3}dx^2.$$

Answer: $d^2y=\frac{2}{(x+2y)^3}dx^2.$

5.315. $x=y-a\sin y.$

Solution.

Let's find $y''(x)$ as the derivative of the function defined implicitly:

$F(x,y)=x-y+a\sin y=0;$

$F'(x,y)=(x-y+a\sin y)'=1-y'+a\cos y y'=0\Rightarrow y'=\frac{1}{1-a\cos y};$

$$F''(x,y)=(1-y'+ay'\cos y)'=-y''+ay''\cos y-ay'\sin y y'=$$ $$=y''(a\cos y-1)-a\left(\frac{1}{1-a\cos y}\right)^2\sin y=0\Rightarrow $$ $$\Rightarrow y''=\frac{a\sin y\left(\frac{1}{1-a\cos y}\right)^2}{a\cos y-1}=\frac{-a\sin y}{(1-a\cos y)^3}.$$

Thus,

$$d^2y=y''(x)dx^2=\frac{-a\sin y}{(1-a\cos y)^3}dx^2.$$

Answer: $d^2y=\frac{-a\sin y}{(1-a\cos y)^3}dx^2.$

Homework:

Find the second-order differentials of the specified functions:

5.304. $y=3^{-x^2}.$

Answer: $3^{-x^2}\ln 9(2x^2\ln 3-1)dx^2$

5.306. $y=ax^2+bx+c.$

Answer: $2adx^2$

5.308. $y=\sqrt{1-x^2}\arcsin x.$

Answer: $-\frac{\sqrt{1-x^2}\cdot x+\arcsin x }{(1-x^2)^{3/2}}dx^2$

5.310. $y=\arcsin(a\sin x).$