The differential of a function. First-order differentials.

Definition. A function $y=f(x)$ is called differentiable at the point $x_0$ if its increment $\mathrm{\Delta }y(x_0,\mathrm{\Delta }x)$ can be represented as $$\mathrm{\Delta }y(x_0,\mathrm{\Delta }x)=A\mathrm{\Delta }x+o(\mathrm{\Delta }x).$$

The main linear part $A\mathrm{\Delta }x$ of the increment $\mathrm{\Delta }y$ is called the differential of this function at the point $x_0$, corresponding to the increment $\mathrm{\Delta }x$, and is denoted by the symbol $dy(x_0,\mathrm{\Delta }x)$.

For the function $y=f(x)$ to be differentiable at the point $x_0$, it is necessary and sufficient for the derivative $f^{\prime }(x_0)$ to exist, and in this case, the equality $A=f^{\prime }(x_0)$ holds.

The expression for the differential is given by:

$$dy(x_0,dx)=f^{\prime }(x_0)dx,$$ where $dx=\mathrm{\Delta }x.$

Properties of the differential:

1. $d(C)=0$, where $C$ is a constant;

2. $d(C_{1}u+C_{2}v)=C_{1}du+C_{2}dv;$

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:

$$dz(x,dx)=z^{\prime }(y)dy(x,dx),$$ 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 $dz(x,dx)=z^{\prime }(x)dx$. 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 $\mathrm{\Delta }x=dx$:

1. $x\sqrt{a^2-x^2}+a^2\arcsin \frac{x}{a}-5.$

Solution.

Let $y(x)=x\sqrt{a^2-x^2}+a^2\arcsin \frac{x}{a}-5$. Then $dy=y^{\prime }(x)dx$.

Let's find $y^{\prime }(x)$:

$y^{\prime }(x)=(x\sqrt{a^2-x^2}+a^2\arcsin \frac{x}{a}-5{)}^{\prime }=$ $=x^{\prime }\sqrt{a^2-x^2}+x(\sqrt{a^2-x^2}{)}^{\prime }+a^2(\arcsin \frac{x}{a}{)}^{\prime }=$ $=\sqrt{a^2-x^2}+\frac{x}{2\sqrt{a^2-x^2}}(a^2-x^2{)}^{\prime }+\frac{a^2}{\sqrt{1-{\left(\frac{x}{a}\right)}^2}}{\left(\frac{x}{a}\right)}^{\prime }=$ $=\sqrt{a^2-x^2}+\frac{x(-2x)}{2\sqrt{a^2-x^2}}+\frac{a^2}{\sqrt{1-{\left(\frac{x}{a}\right)}^2}}\left(\frac{1}{a}\right)=$ $=\sqrt{a^2-x^2}-\frac{x^2}{\sqrt{a^2-x^2}}+\frac{a^2}{\frac{1}{a}\sqrt{a^2-x^2}}\left(\frac{1}{a}\right)=\frac{a^2-x^2-x^2+a^2}{\sqrt{a^2-x^2}}=2\sqrt{a^2-x^2}.$

Thus, $dy=2\sqrt{a^2-x^2}dx$.

Answer: $dy=2\sqrt{a^2-x^2}dx$.

2. $\sin x-x\cos x+4$.

Solution.

Let $y(x)=\sin x-x\cos x+4$.

Then $dy=y^{\prime }(x)dx$.

Let's find $y^{\prime }(x)$:

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

Thus, $dy=x\sin xdx$.

Answer: $dy=x\sin xdx$.

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

3. $y^5+y-x^2=1$.

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:

$d(y^5+y-x^2)=d(y^5)+dy-d(x^2)=5y^{4}dy+dy-2xdx=$ $=-2xdx+(5y^4+1)dy;$

$d(1)=0.$

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: $dy=\frac{2x}{5y^4+1}dx.$

4. $e^y=x+y.$

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:

$d(e^y)=e^{y}dy;$

$d(x+y)=dx+dy.$

By equating the obtained expressions, we get $e^{y}dy=dx+dy$. 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. $\cos (xy)=x.$

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 $\mathrm{\Delta }x=dx$:

1. $xarctgx-\ln \sqrt{1+x^2}.$

Answer: $arctgxdx.$

2. $x\ln x-x+1.$

Answer: $lnxdx.$

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

3. $x^4+y^4=x^2{y}^2.$

Answer: $\frac{x(y^2-2x^2)}{y(2y^2-x^2)}dx.$

4. $y=x+arctgy.$

Answer: $\frac{y^2-1}{y^2}dx.$

5. $y=\cos (x+y).$

Answer: $\frac{\sin (x+y)}{1+\sin (x+y)}dx.$

6. $arctg\frac{y}{x}=\ln \sqrt{x^2+y^2}.$

Answer: $\frac{x+y}{x-y}dx.$

Tags: calculus, derivative, differential, masematical analysis