Differentiation of complex and implicitly defined functions.

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

Complex functions of one and several independent variables.

If $u = f(x_1, x_2, .., x_n)$ is a differentiable function of variables $x_1, x_2, ..., x_n$, which themselves are differentiable functions of the independent variable $t$:

$$x_1=\varphi_1(t),\quad x_2=\varphi_2(t),\quad, x_n=\varphi_n(t),$$ The derivative of the composite function $u = f(\varphi_1(t), \varphi_2(t), ..., \varphi_n(t))$ is calculated by the formula: $$\frac{du}{dt}=\frac{\partial u}{\partial x_1}.\frac{dx_1}{dt}+\frac{\partial u}{\partial x_2}.\frac{dx_2}{dt}+...+\frac{\partial u}{\partial x_n}.\frac{dx_n}{dt}.$$ In particular, if $t$ coincides, for example, with the variable $x_1$, then the "total" derivative of the function $u$ with respect to $x_1$ is equal to:

$$\frac{du}{dx_1}=\frac{\partial u}{\partial x_1}+\frac{\partial u}{\partial x_2}\cdot\frac{dx_2}{dx_1}+...+\frac{\partial u}{\partial x_n}\cdot\frac{dx_n}{dx_1}.$$ Let $u = f(x_1, x_2, .., x_n),$ where $$x_1=\varphi_1(t_1, t_2, ..., t_m),\quad x_2=\varphi_2(t_1, t_2, ..., t_m),\quad, x_n=\varphi_n(t_1, t_2, ..., t_m),$$ $(t_1, t_2,..., t_m)$ - independent variables. The partial derivatives of the function $u$ with respect to $t_1, t_2, ..., t_m$ are expressed as follows:$$\frac{\partial u}{\partial t_1}=\frac{\partial u}{\partial x_1}\cdot\frac{\partial x_1}{\partial t_1}+\frac{\partial u}{\partial x_2}\cdot\frac{\partial x_2}{\partial t_1}+...+\frac{\partial u}{\partial x_n}\cdot\frac{\partial x_n}{\partial t_1},$$ $$\frac{\partial u}{\partial t_2}=\frac{\partial u}{\partial x_1}\cdot\frac{\partial x_1}{\partial t_2}+\frac{\partial u}{\partial x_2}\cdot\frac{\partial x_2}{\partial t_2}+...+\frac{\partial u}{\partial x_n}\cdot\frac{\partial x_n}{\partial t_2},$$ $$\cdots$$ $$\frac{\partial u}{\partial t_m}=\frac{\partial u}{\partial x_1}\cdot\frac{\partial x_1}{\partial t_m}+\frac{\partial u}{\partial x_2}\cdot\frac{\partial x_2}{\partial t_m}+...+\frac{\partial u}{\partial x_n}\cdot\frac{\partial x_n}{\partial t_m}.$$In this case, the expression for the first-order differential remains unchanged. $$du=\frac{\partial u}{\partial x_1}dx_1+\frac{\partial u}{\partial x_2}dx_2+...+\frac{\partial u}{\partial x_n}dx_n.$$ The expressions for higher-order differentials of a composite function, generally speaking, differ from the expression of the form $$d^mu=\left(\frac{\partial}{\partial x_1}dx_1+\frac{\partial}{\partial x_2}dx_2+...+\frac{\partial}{\partial x_n}dx_n\right)^mu.$$ For example, the second-order differential is expressed by the formula

$$d^2u=\left(\frac{\partial}{\partial x_1}dx_1+\frac{\partial}{\partial x_2}dx_2+...+\frac{\partial}{\partial x_n}dx_n\right)^2u+$$ $$+\frac{\partial u}{\partial x_1}d^2x_1+\frac{\partial u}{\partial x_1}d^2x_2+...+\frac{\partial u}{\partial x_n} d^2x_n.$$

Implicit functions of one and several independent variables.

Let the equation $f(x, y) = 0$, where $f$ is a differentiable function of variables $x$ and $y$, define $y$ as a function of $x$. The first derivative of this implicit function $y = y(x)$ at the point $x_0$ is expressed by the formula:$$\left.\frac{dy}{dx}\right|_{x_0}=-\frac{f'_x(x_0, y_0)}{f'_y(x_0, y_0)}\qquad\qquad\qquad(1)$$ provided that $f'_y(x_0, y_0)\neq 0$, where $y_0=y(x_0)$, $f(x_0, y_0)=0$.

Higher-order derivatives are computed by successive differentiation of formula (1).

Examples:

7.114. Find $\frac{dz}{dt}$ if $z=e^{2x-3y}$, where $x=\tan t$, $y=t^2-t$.

Solution.

We will use the formula

$$\frac{du}{dt}=\frac{\partial u}{\partial x_1}.\frac{dx_1}{dt}+\frac{\partial u}{\partial x_2}.\frac{dx_2}{dt}+...+\frac{\partial u}{\partial x_n}.\frac{dx_n}{dt}.$$

Let's find the partial derivatives:

$\frac{\partial z}{\partial x}=e^{2x-3y}(2x-3y)'_x=2e^{2x-3y};$

$\frac{\partial z}{\partial y}=e^{2x-3y}(2x-3y)'_y=-3e^{2x-3y};$

$\frac{dx}{dt}=\frac{1}{\cos^2t};$

$\frac{dy}{dt}=2t-1.$

Hence,

$$\frac{dz}{dt}=2e^{2x-3y}\frac{1}{\cos^2t}-3e^{2x-3y}(2t-1)=\frac{2e^{2tg t-3(t^2-2)}}{\cos^2 t}-3e^{(2tg t-3t^2-2)}(2t-1).$$

Answer: $\frac{2e^{2tg t-3(t^2-2)}}{\cos^2t}-3e^{3(t^2-2)}(2t-1).$

7.115. Find $\frac{dz}{dt}$ if $z=x^y$, where $x=\ln t$, $y=\sin t$.

Solution.

We will use the formula $$\frac{du}{dt}=\frac{\partial u}{\partial x_1}.\frac{dx_1}{dt}+\frac{\partial u}{\partial x_2}.\frac{dx_2}{dt}+...+\frac{\partial u}{\partial x_n}.\frac{dx_n}{dt}.$$

Let's find the partial derivatives:

$\frac{\partial z}{\partial x}=(x^y)'_x=yx^{y-1};$

$\frac{\partial z}{\partial y}=(x^y)'_y=x^y\ln x;$

$\frac{dx}{dt}=(\ln t)'=\frac{1}{t};$

$\frac{dy}{dt}=(\sin t)'=\cos t.$

Hence

$$\frac{dz}{dt}=yx^{y-1}\frac{1}{t} +x^y \ln x\cos t=\frac{\sin t\cdot \ln t^{\sin t -1}}{t}-(\ln t)^{\sin t}\cos t\ln\ln t.$$

Answer: $\frac{dz}{dt}=\frac{\sin t\cdot \ln t^{\sin t -1}}{t}-(\ln t)^{\sin t}\cos t\ln\ln t.$

7.118. Find $\frac{\partial z}{\partial x}$ and $\frac{dz}{dx},$ if $z=\ln(e^x+e^y),$ where $ y=\frac{1}{3}x^3+x.$

Solution.

$$\frac{\partial z}{\partial x}=(\ln(e^x+e^y))'_x=\frac{1}{e^x+e^y}(e^x+e^y)'_x=\frac{e^x}{e^x+e^y}.$$

To find $\frac{dz}{dx}$, we will use the formula:$$\frac{du}{dx_1}=\frac{\partial u}{\partial x_1}+\frac{\partial u}{\partial x_2}\cdot\frac{dx_2}{dx_1}+...+\frac{\partial u}{\partial x_n}\cdot\frac{dx_n}{dx_1}.$$

$$\frac{\partial z}{\partial y}=(\ln(e^x+e^y))'_y=\frac{1}{e^x+e^y}(e^x+e^y)'_y=\frac{e^y}{e^x+e^y};$$

$$\frac{dy}{dx}=(\frac{1}{3}x^3+x)'=\frac{3x^2}{3}+1=x^2+1.$$

Hence

$$\frac{dz}{dz}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot\frac{dy}{dx}=\frac{e^x}{e^x+e^y}+\frac{e^y}{e^x+e^y}(x^2+1).$$

Answer: $\frac{e^x}{e^x+e^y};$ $\frac{e^x+e^y(x^2+1)}{e^x+e^y}.$

7.123. Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y},$ if $z=f(u, v),$ where $ u=\ln(x^2-y^2),\, v=xy^2.$

Solution.

We will use the formulas$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x},$$ $$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial y},$$

Let's find the partial derivatives:

$$\frac{\partial u}{\partial x}=(\ln(x^2-y^2))'_x=\frac{1}{x^2-y^2}(x^2-y^2)'_x=\frac{2x}{x^2-y^2};$$

$$\frac{\partial u}{\partial y}=(\ln(x^2-y^2))'_y=\frac{1}{x^2-y^2}(x^2-y^2)'_y=\frac{-2y}{x^2-y^2};$$

$$\frac{\partial v}{\partial x}=(xy^2)'_x=y^2;$$

$$\frac{\partial v}{\partial y}=(xy^2)'_y=2xy;$$

Hence

$$\frac{\partial z}{\partial x}=z'_u\frac{2x}{x^2-y^2}+z'_vy^2,$$

$$\frac{\partial z}{\partial y}=z'_u\frac{-2y}{x^2-y^2}+2z'_vxy.$$

Answer: $\frac{\partial z}{\partial x}=z'_u\frac{2x}{x^2-y^2}+z'_vy^2,$ $\frac{\partial z}{\partial y}=z'_u\frac{-2y}{x^2-y^2}+2z'_vxy.$

7.125. Find $dz,$ if $z=f(u, v),$ where $ u=\sin\frac{x}{y},\, v=\sqrt{x/y}.$

Solution.

We will use the formula $$dz=\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial y}dy.$$

Let's find the partial derivatives:

$$\frac{\partial u}{\partial x}=\left(\sin\frac{x}{y}\right)'_x=\frac{1}{y}\cos\frac{x}{y};$$

$$\frac{\partial u}{\partial y}=\left(\sin\frac{x}{y}\right)'_y=\frac{-x}{y^2}\cos\frac{x}{y};$$

$$\frac{\partial v}{\partial x}=\left(\sqrt{x/y}\right)'_x=\frac{1}{2\sqrt{x/y}}\left(\frac{x}{y}\right)'_x=\frac{1}{2y\sqrt{x/y}};$$

$$\frac{\partial v}{\partial y}=\left(\sqrt{x/y}\right)'_y=\frac{1}{2\sqrt{x/y}}\left(\frac{x}{y}\right)'_y=\frac{-x}{2y^2\sqrt{x/y}}.$$

Hence

$$du=\frac{1}{y}\cos\frac{x}{y}dx-\frac{x}{y^2}\cos\frac{x}{y}dy.$$

$$dv=\frac{1}{2y\sqrt{x/y}}dx-\frac{x}{2y^2\sqrt{x/y}}dy.$$

Thus,

$$dz=f'_u\left(\frac{1}{y}\cos\frac{x}{y}dx-\frac{x}{y^2}\cos\frac{x}{y}dy\right)+f'_v\left(\frac{1}{2y\sqrt{x/y}}dx-\frac{x}{2y^2\sqrt{x/y}}dy\right)=$$ $$=\frac{1}{y^2}\left(yf'_u\cos\frac{x}{y}+yf'_v\frac{1}{2\sqrt{x/y}}\right)dx-\left(x\cos\frac{x}{y}+\frac{x}{2\sqrt{x/y}}\right)dy.$$

Answer: $\frac{1}{y^2}\left(yf'_u\cos\frac{x}{y}+yf'_v\frac{1}{2\sqrt{x/y}}\right)dx-\left(x\cos\frac{x}{y}+\frac{x}{2\sqrt{x/y}}\right)dy.$

7.138. Find $d^2u,$ if $u=f(ax,by,cz).$

Solution.

Let's denote $$x_1=ax,$$ $$x_2=by,$$ $$x_3=cz.$$ We will use the formula

$$d^2u=\left(\frac{\partial}{\partial x_1}dx_1+ \frac{\partial}{\partial x_2}dx_2+\frac{\partial}{\partial x_3}dx_3\right)^2u+\frac{\partial u}{\partial x_1}d^2x_1+ \frac{\partial u}{\partial x_1}d^2x_2+\frac{\partial u}{\partial x_3} d^2x_3.$$

Next, we find,

$$dx_1=d(ax)=adx;$$

$$dx_2=d(by)=bdy;$$

$$dx_3=d(cz)=cdz;$$

$$d^2x_1=(ax)''_xdx^2=0;$$

$$d^2x_2=(by)''_ydy^2=0;$$

$$d^2x_3=(cz)''_zdz^2=0.$$

Thus,

$$d^2u=\left(\frac{\partial}{\partial x_1}dx_1+ \frac{\partial}{\partial x_2}dx_2+\frac{\partial}{\partial x_3}dx_3\right)^2u+\frac{\partial u}{\partial x_1}d^2x_1+ \frac{\partial u}{\partial x_1}d^2x_2+\frac{\partial u}{\partial x_3} d^2x_3=$$

$$=\frac{\partial^2u}{\partial x_1^2}dx_1^2+ \frac{\partial^2u}{\partial x_2^2}dx_2^2+\frac{\partial^2u}{\partial x_3^2}dx_3^2+2\frac{\partial^2u}{\partial x_1\partial x_2}dx_1dx_2+ 2\frac{\partial^2u}{\partial x_1x_3}dx_1dx_3+2\frac{\partial^2u}{\partial x_2\partial x_3}dx_2dx_3=$$ $$=\frac{\partial^2u}{\partial x_1^2}a^2dx^2+ \frac{\partial^2u}{\partial x_2^2}b^2dy^2+\frac{\partial^2u}{\partial x_3^2}c^2dz^2+2\frac{\partial^2u}{\partial x_1\partial x_2}abdxdy+ 2\frac{\partial^2u}{\partial x_1x_3}acdxdz+2\frac{\partial^2u}{\partial x_2\partial x_3}bcdydz.$$

Answer: $\frac{\partial^2u}{\partial x_1^2}a^2dx^2+ \frac{\partial^2u}{\partial x_2^2}b^2dy^2+\frac{\partial^2u}{\partial x_3^2}c^2dz^2+2\frac{\partial^2u}{\partial x_1\partial x_2}abdxdy+ 2\frac{\partial^2u}{\partial x_1x_3}acdxdz+2\frac{\partial^2u}{\partial x_2\partial x_3}bcdydz.$

7.140. Find $\frac{dy}{dx},$ if $x^2e^{2y}-y^2e^{2x}=0.$

Solution.

We find the derivative $\frac{dy}{dx}$ using the formula: $$\frac{dy}{dx}=-\frac{f'_x(x, y)}{f'_y(x, y)}.$$ Here $f(x,y)=x^2e^{2y}-y^2e^{2x}.$

Let's find the partial derivatives:

$$f'_x(x,y)=(x^2e^{2y}-y^2e^{2x})'_x=2xe^{2y}-2y^2e^{2x};$$

$$f'_y(x,y)=(x^2e^{2y}-y^2e^{2x})'_y=2x^2e^{2y}-2ye^{2x}.$$

From here, we find

$$\frac{dy}{dx}=-\frac{f'_x(x, y)}{f'_y(x, y)}=-\frac{2xe^{2y}-2y^2e^{2x}}{2x^2e^{2y}-2ye^{2x}}=\frac{y^2e^{2x}-xe^{2y}}{x^2e^{2y}-ye^{2x}}.$$

Answer: $\frac{y^2e^{2x}-xe^{2y}}{x^2e^{2y}-ye^{2x}}.$

7.143. Find $\frac{dy}{dx},$ $\frac{d^2y}{dx^2},$ if $x-y+arctg y=0.$

Solution.

We find the derivative $\frac{dy}{dx}$ using the formula$$\frac{dy}{dx}=-\frac{f'_x(x, y)}{f'_y(x, y)}.$$ Here $f(x,y)=x-y+arctg y.$

Let's find the partial derivatives:

$$f'_x(x,y)=(x-y+arctgy)'_x=1;$$

$$f'_y(x,y)=(x-y+arctgy)'_y=-1+\frac{1}{1+y^2}.$$

From here, we find

$$\frac{dy}{dx}=-\frac{f'_x(x, y)}{f'_y(x, y)}=-\frac{1}{-1+\frac{1}{1+y^2}}=\frac{1+y^2}{y^2}.$$

We find the second-order derivative $\frac{d^2y}{dx^2}$ by differentiating the expression $\frac{dy}{dx}=\frac{1+y^2}{y^2}$ with respect to the variable $x$.

$$\frac{d^2y}{dx^2}=\frac{(1+y^2)'_xy^2-(y^2)'_x(1+y^2)}{y^4}=\frac{2yy'_xy^2-2yy'_x(1+y^2)}{y^4}=-2\frac{y'_x}{y^3}=$$ $$=-2\frac{\frac{1+y^2}{y^2}}{y^3}=-2\frac{1+y^2}{y^5}$$

Answer: $\frac{dy}{dx}=\frac{1+y^2}{y^2};$ $\frac{d^2y}{dx^2}=-2\frac{1+y^2}{y^5}.$

7.152. Find $\frac{\partial^2z}{\partial x^2},$ $\frac{\partial^2z}{\partial x\partial y},$ $\frac{\partial^2z}{\partial y^2},$ if $x+y+z=e^z.$

Solution.

We find the derivatives $\frac{dz}{dx}$ and $\frac{dz}{dy}$ using the formulas$$\frac{dz}{dx}=-\frac{f'_x(x, y, z)}{f'_z(x, y, z)};$$ $$\frac{dz}{dy}=-\frac{f'_y(x, y, z)}{f'_z(x, y, z)};$$ Here $f(x, y, z)=x+y+z-e^z.$

Let's find the partial derivatives:

$$f'_x(x,y,z)=(x+y+z-e^z)'_x=1;$$

$$f'_y(x,y,z)=(x+y+z-e^z)'_y=1;$$

$$f'_z(x,y,z)=(x+y+z-e^z)'_z=1-e^z.$$

From here, we find

$$\frac{dz}{dx}=-\frac{f'_x(x, y, z)}{f'_z(x, y, z)}=-\frac{1}{1-e^z}.$$

$$\frac{dz}{dy}=-\frac{f'_x(x, y, z)}{f'_z(x, y, z)}=-\frac{1}{1-e^z}.$$

Second-order derivatives are found by differentiating the found first-order derivatives with respect to the corresponding variables.

$$\frac{d^2z}{dx^2}=\frac{(1-e^z)'_x}{(1-e^z)^2}=-\frac{e^zz'_x}{(1-e^z)^2}=-\frac{-e^z\cdot\frac{1}{1-e^z}}{(1-e^z)^2}=\frac{e^z}{(1-e^z)^3}.$$ $$\frac{d^2z}{dxdy}=\frac{(1-e^z)'_y}{(1-e^z)^2}=-\frac{e^zz'_y}{(1-e^z)^2}=-\frac{-e^z\cdot\frac{1}{1-e^z}}{(1-e^z)^2}=\frac{e^z}{(1-e^z)^3}.$$ $$\frac{d^2z}{dy^2}=\frac{(1-e^z)'_y}{(1-e^z)^2}=-\frac{e^zz'_y}{(1-e^z)^2}=-\frac{-e^z\cdot\frac{1}{1-e^z}}{(1-e^z)^2}=\frac{e^z}{(1-e^z)^3}.$$

Answer: $\frac{d^2z}{dx^2}=\frac{d^2z}{dxdy}=\frac{d^2z}{dy^2}=\frac{e^z}{(1-e^z)^3}.$