The differential of a function. First-order differentials.
Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.
Definition. A function $y=f(x)$ is called differentiable at the point $x_0$ if its increment $\Delta y(x_0, \Delta x)$ can be represented as $$\Delta y(x_0, \Delta x)=A\Delta x+o(\Delta x).$$
The main linear part $A\Delta x$ of the increment $\Delta y$ is called the differential of this function at the point $x_0$, corresponding to the increment $\Delta x$, and is denoted by the symbol $dy(x_0, \Delta x)$.
For the function $y=f(x)$ to be differentiable at the point $x_0$, it is necessary and sufficient for the derivative $f'(x_0)$ to exist, and in this case, the equality $A=f'(x_0)$ holds.
The expression for the differential is given by:
$$dy(x_0, dx)=f'(x_0)dx,$$ where $dx=\Delta x.$
Properties of the differential:
1. $d(C)=0$, where $C$ is a constant;
2. $d(C_1u+C_2v)=C_1du+C_2dv;$
3. $d(uv)=udv+vdu;$
4. $d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2};$
5. Let $z(x)=z(y(x))$ be a composite function formed by the composition of functions $y=y(x)$ and $z=z(y)$. Then:
$$dz(x, dx)=z'(y)dy(x, dx), $$ In other words, the expression for the differential of a composite function through the differential of the intermediate argument has the same form as the main definition $dz(x, dx)=z'(x)dx$. This statement is called the invariance of the form of the 1st differential.
Examples.
Find the differentials of the specified functions for arbitrary values of the argument $x$ and for arbitrary increments $\Delta x=dx$:
5.285. $x\sqrt{a^2-x^2}+a^2\arcsin\frac{x}{a}-5.$
Solution.
Let $y(x)=x\sqrt{a^2-x^2}+a^2\arcsin\frac{x}{a}-5$. Then $dy=y'(x)dx$.
Let's find $y'(x)$:
$y'(x)=(x\sqrt{a^2-x^2}+a^2\arcsin\frac{x}{a}-5)'=$ $=x'\sqrt{a^2-x^2}+x(\sqrt{a^2-x^2})'+a^2(\arcsin\frac{x}{a})'=$ $=\sqrt{a^2-x^2}+\frac{x}{2\sqrt{a^2-x^2}}(a^2-x^2)'+\frac{a^2}{\sqrt{1-\left(\frac{x}{a}\right)^2}}\left(\frac{x}{a}\right)'=$ $=\sqrt{a^2-x^2}+\frac{x(-2x)}{2\sqrt{a^2-x^2}}+\frac{a^2}{\sqrt{1-\left(\frac{x}{a}\right)^2}}\left(\frac{1}{a}\right)=$ $=\sqrt{a^2-x^2}-\frac{x^2}{\sqrt{a^2-x^2}}+\frac{a^2}{\frac{1}{a}\sqrt{a^2-x^2}}\left(\frac{1}{a}\right)=\frac{a^2-x^2-x^2+a^2}{\sqrt{a^2-x^2}}=2\sqrt{a^2-x^2}.$
Thus, $dy=2\sqrt{a^2-x^2}dx$.
Answer: $dy=2\sqrt{a^2-x^2}dx$.
5.286. $\sin x-x\cos x+4$.
Solution.
Let $y(x)=\sin x-x\cos x+4$.
Then $dy=y'(x)dx$.
Let's find $y'(x)$:
$y'(x)=(\sin x-x\cos x+4)'=\cos x-x'\cos x-x(\cos x)'+4'=$ $=\cos x-\cos x+x\sin x=x\sin x$.
Thus, $dy=x\sin xdx$.
Answer: $dy=x\sin xdx$.
Find the differentials of the following functions implicitly defined $y=y(x)$:
5.290. $y^5+y-x^2=1$.
Solution.
Rewrite the given equation as an identity:
$y^5(x)+y(x)-x^2=1$
and compute the differentials of the left and right sides. Using the properties of the differential, we find:
$d(y^5+y-x^2)=d(y^5)+dy-d(x^2)=5y^4dy+dy-2xdx=$ $=-2xdx+(5y^4+1)dy;$
$d(1)=0.$
By equating the obtained expressions, we get $-2xdx+(5y^4+1)dy=0$. From this equation, we express $dy$ in terms of $x$, $y$, and $dx$:
$dy=\frac{2x}{5y^4+1}dx.$
Answer: $dy=\frac{2x}{5y^4+1}dx.$
5.293. $e^y=x+y.$
Solution.
Rewrite the given equation as an identity:
$e^y(x)=x+y(x).$
and compute the differentials of the left and right sides. Using the properties of the differential, we find:
$d(e^y)=e^ydy;$
$d(x+y)=dx+dy.$
By equating the obtained expressions, we get $e^ydy=dx+dy$. From this equation, we express $dy$ in terms of $x$, $y$, and $dx$:
$(e^y-1)dy=dx\Rightarrow dy=\frac{1}{e^y-1}dx$
Answer: $ dy=\frac{1}{e^y-1}dx$
5. 297. $\cos (xy)=x.$
Solution.
Let's rewrite the given equation as an identity:
$\cos (xy(x))=x.$
and compute the differentials of the left and right sides. Using the properties of the differential, we find:
$d(\cos (xy))=-\sin (xy)d(xy)=-\sin (xy)(ydx+xdy)=-y\sin (xy)dx-x\sin (xy)dy;$
$d(x)=dx.$
By equating the obtained expressions, we get $-y\sin (xy)dx-x\sin (xy)dy=dx$. From this equation, we express $dy$ in terms of $x$, $y$, and $dx$:
$x\sin (xy)dy=-(1+y\sin (xy))dx\Rightarrow dy=-\frac{1+y\sin (xy)}{x\sin (xy)}dx$
Answer: $dy=-\frac{1+y\sin (xy)}{x\sin (xy)}dx$
Homework.
Find the differentials of the specified functions for arbitrary values of the argument $x$ and for arbitrary increments $\Delta x=dx$:
5.287. $x arctg x-\ln\sqrt{1+x^2}.$
Answer: $arctg xdx.$
5.288. $x\ln x-x+1.$
Answer: $ln xdx.$
Find the differentials of the following implicitly defined functions $y=y(x)$:
5.291. $x^4+y^4=x^2y^2.$
Answer: $\frac{x(y^2-2x^2)}{y(2y^2-x^2)}dx.$
5.294. $y=x+arctg y.$
Answer: $\frac{y^2-1}{y^2}dx.$
5.295. $y=\cos (x+y).$
Answer: $\frac{\sin(x+y)}{1+\sin(x+y)}dx.$
5.296. $arctg\frac{y}{x}=\ln\sqrt{x^2+y^2}.$
Answer: $\frac{x+y}{x-y}dx.$
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