Calculation of higher-order differentials.

Let's consider the differential $dy(x,dx)=f^{\prime }(x)dx$ 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 $d^{2}y(x,dx)$.

Thus, $$d^{2}y=f^{″}(x)d{x}^2$$

Similarly,

$$d^{3}y=d(d^{2}y)=f^{‴}(x)d{x}^3$$

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

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

Examples:

Find the second-order differentials of the specified functions:

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

Solution.

$y^{\prime }(x)=(a\sin (bx+c){)}^{\prime }=a\cos (bx+c)(bx+c{)}^{\prime }=ab\cos (bx+c);$

$y^{″}(x)=(ab\cos (bx+c){)}^{\prime }=-ab\sin (bx+c)(bx+c{)}^{\prime }=-a{b}^2\sin (bx+c).$

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

$$d^{2}y=f^{″}(x)d{x}^2.$$

Thus, we obtain

$$d^{2}y=-a{b}^2\sin (bx+c)d{x}^2.$$

Answer: $d^{2}y=-a{b}^2\sin (bx+c)d{x}^2.$

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

Solution.

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

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

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

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

$$d^{2}y=f^{″}(x)d{x}^2.$$

Thus, we obtain

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

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

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

Solution.

Let's find $y^{″}:$

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

$$y^{″}=(-(x^2-3x+2{)}^{-2}(2x-3){)}^{\prime }=$$ $$=-((x^2-3x+2{)}^{-2}{)}^{\prime }(2x-3)-(x^2-3x+2{)}^{-2}(2x-3{)}^{\prime }=$$

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

We find the second-order differential using the formula

$$d^{2}y=f^{″}(x)d{x}^2.$$

Thus, we obtain

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

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

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

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

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 $d^{2}y:$

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

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

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

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

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

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

Solution.

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

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

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

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

Thus,

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

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

Homework:

Find the second-order differentials of the specified functions:

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

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

2. $y=a{x}^2+bx+c.$

Answer: $2ad{x}^2$

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

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

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

Answer: $-\frac{a(1+a^2)\sin x}{(1+a^2{\sin }^{2}x{)}^{3/2}}d{x}^2$

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

5. $(x-a{)}^2+(y-b{)}^2=R^2.$

Answer: $-\frac{R^2}{(y-b{)}^3}d{x}^2$

6. $x^3+y^3=y.$

Answer: $6\frac{x(1+3y^2)}{(1-3y^2{)}^3}d{x}^2.$

Tags: calculus, differential, higher-order differentials, mathematical analysis