Differentiation of complex and implicitly defined functions.

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 $t$:

$$x_1={\phi }_1(t),x_2={\phi }_2(t),,x_n={\phi }_n(t),$$ The derivative of the composite function $u=f({\phi }_1(t),{\phi }_2(t),...,{\phi }_n(t))$ is calculated by the formula: $$\frac{du}{dt}=\frac{\partial u}{\partial x_1}.\frac{d{x}_1}{dt}+\frac{\partial u}{\partial x_2}.\frac{d{x}_2}{dt}+...+\frac{\partial u}{\partial x_n}.\frac{d{x}_n}{dt}.$$ 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:

$$\frac{du}{d{x}_1}=\frac{\partial u}{\partial x_1}+\frac{\partial u}{\partial x_2}\cdot \frac{d{x}_2}{d{x}_1}+...+\frac{\partial u}{\partial x_n}\cdot \frac{d{x}_n}{d{x}_1}.$$ Let $u=f(x_1,x_2,..,x_n),$ where $$x_1={\phi }_1(t_1,t_2,...,t_m),x_2={\phi }_2(t_1,t_2,...,t_m),,x_n={\phi }_n(t_1,t_2,...,t_m),$$ $(t_1,t_2,...,t_m)$ - 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. $$du=\frac{\partial u}{\partial x_1}d{x}_1+\frac{\partial u}{\partial x_2}d{x}_2+...+\frac{\partial u}{\partial x_n}d{x}_n.$$ The expressions for higher-order differentials of a composite function, generally speaking, differ from the expression of the form $$d^{m}u={(\frac{\partial }{\partial x_1}d{x}_1+\frac{\partial }{\partial x_2}d{x}_2+...+\frac{\partial }{\partial x_n}d{x}_n)}^{m}u.$$ For example, the second-order differential is expressed by the formula

$$d^{2}u={(\frac{\partial }{\partial x_1}d{x}_1+\frac{\partial }{\partial x_2}d{x}_2+...+\frac{\partial }{\partial x_n}d{x}_n)}^{2}u+$$ $$+\frac{\partial u}{\partial x_1}{d}^2{x}_1+\frac{\partial u}{\partial x_1}{d}^2{x}_2+...+\frac{\partial u}{\partial x_n}{d}^2{x}_n.$$

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:$${\frac{dy}{dx}|}_{x_0}=-\frac{f_x^{\prime }(x_0,y_0)}{f_y^{\prime }(x_0,y_0)}(1)$$ provided that $f_y^{\prime }(x_0,y_0)\ne 0$, where $y_0=y(x_0)$, $f(x_0,y_0)=0$.

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

Examples:

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

Solution.

We will use the formula

$$\frac{du}{dt}=\frac{\partial u}{\partial x_1}.\frac{d{x}_1}{dt}+\frac{\partial u}{\partial x_2}.\frac{d{x}_2}{dt}+...+\frac{\partial u}{\partial x_n}.\frac{d{x}_n}{dt}.$$

Let's find the partial derivatives:

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

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

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

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

Hence,

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

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

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

Solution.

We will use the formula $$\frac{du}{dt}=\frac{\partial u}{\partial x_1}.\frac{d{x}_1}{dt}+\frac{\partial u}{\partial x_2}.\frac{d{x}_2}{dt}+...+\frac{\partial u}{\partial x_n}.\frac{d{x}_n}{dt}.$$

Let's find the partial derivatives:

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

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

$\frac{dx}{dt}=(\ln t{)}^{\prime }=\frac{1}{t};$

$\frac{dy}{dt}=(\sin t{)}^{\prime }=\cos t.$

Hence

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

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

3. 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.

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

To find $\frac{dz}{dx}$, we will use the formula:$$\frac{du}{d{x}_1}=\frac{\partial u}{\partial x_1}+\frac{\partial u}{\partial x_2}\cdot \frac{d{x}_2}{d{x}_1}+...+\frac{\partial u}{\partial x_n}\cdot \frac{d{x}_n}{d{x}_1}.$$

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

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

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

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

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:

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

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

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

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

Hence

$$\frac{\partial z}{\partial x}=z_u^{\prime }\frac{2x}{x^2-y^2}+z_v^{\prime }{y}^2,$$

$$\frac{\partial z}{\partial y}=z_u^{\prime }\frac{-2y}{x^2-y^2}+2z_v^{\prime }xy.$$

Answer: $\frac{\partial z}{\partial x}=z_u^{\prime }\frac{2x}{x^2-y^2}+z_v^{\prime }{y}^2,$ $\frac{\partial z}{\partial y}=z_u^{\prime }\frac{-2y}{x^2-y^2}+2z_v^{\prime }xy.$

5. 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:

$$\frac{\partial u}{\partial x}={(\sin \frac{x}{y})}_x^{\prime }=\frac{1}{y}\cos \frac{x}{y};$$

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

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

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

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,

$$dz=f_u^{\prime }(\frac{1}{y}\cos \frac{x}{y}dx-\frac{x}{y^2}\cos \frac{x}{y}dy)+f_v^{\prime }(\frac{1}{2y\sqrt{x/y}}dx-\frac{x}{2y^2\sqrt{x/y}}dy)=$$ $$=\frac{1}{y^2}(y{f}_u^{\prime }\cos \frac{x}{y}+y{f}_v^{\prime }\frac{1}{2\sqrt{x/y}})dx-(x\cos \frac{x}{y}+\frac{x}{2\sqrt{x/y}})dy.$$

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

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

Solution.

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

$$d^{2}u={(\frac{\partial }{\partial x_1}d{x}_1+\frac{\partial }{\partial x_2}d{x}_2+\frac{\partial }{\partial x_3}d{x}_3)}^{2}u+\frac{\partial u}{\partial x_1}{d}^2{x}_1+\frac{\partial u}{\partial x_1}{d}^2{x}_2+\frac{\partial u}{\partial x_3}{d}^2{x}_3.$$

Next, we find,

$$d{x}_1=d(ax)=adx;$$

$$d{x}_2=d(by)=bdy;$$

$$d{x}_3=d(cz)=cdz;$$

$$d^2{x}_1=(ax{)}_x^{″}d{x}^2=0;$$

$$d^2{x}_2=(by{)}_y^{″}d{y}^2=0;$$

$$d^2{x}_3=(cz{)}_z^{″}d{z}^2=0.$$

Thus,

$$d^{2}u={(\frac{\partial }{\partial x_1}d{x}_1+\frac{\partial }{\partial x_2}d{x}_2+\frac{\partial }{\partial x_3}d{x}_3)}^{2}u+\frac{\partial u}{\partial x_1}{d}^2{x}_1+\frac{\partial u}{\partial x_1}{d}^2{x}_2+\frac{\partial u}{\partial x_3}{d}^2{x}_3=$$

$$=\frac{{\partial }^{2}u}{\partial x_1^2}d{x}_1^2+\frac{{\partial }^{2}u}{\partial x_2^2}d{x}_2^2+\frac{{\partial }^{2}u}{\partial x_3^2}d{x}_3^2+2\frac{{\partial }^{2}u}{\partial x_1\partial x_2}d{x}_{1}d{x}_2+2\frac{{\partial }^{2}u}{\partial x_1{x}_3}d{x}_{1}d{x}_3+2\frac{{\partial }^{2}u}{\partial x_2\partial x_3}d{x}_{2}d{x}_3=$$ $$=\frac{{\partial }^{2}u}{\partial x_1^2}{a}^{2}d{x}^2+\frac{{\partial }^{2}u}{\partial x_2^2}{b}^{2}d{y}^2+\frac{{\partial }^{2}u}{\partial x_3^2}{c}^{2}d{z}^2+2\frac{{\partial }^{2}u}{\partial x_1\partial x_2}abdxdy+2\frac{{\partial }^{2}u}{\partial x_1{x}_3}acdxdz+2\frac{{\partial }^{2}u}{\partial x_2\partial x_3}bcdydz.$$

Answer: $\frac{{\partial }^{2}u}{\partial x_1^2}{a}^{2}d{x}^2+\frac{{\partial }^{2}u}{\partial x_2^2}{b}^{2}d{y}^2+\frac{{\partial }^{2}u}{\partial x_3^2}{c}^{2}d{z}^2+2\frac{{\partial }^{2}u}{\partial x_1\partial x_2}abdxdy+2\frac{{\partial }^{2}u}{\partial x_1{x}_3}acdxdz+2\frac{{\partial }^{2}u}{\partial x_2\partial x_3}bcdydz.$

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

Solution.

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

Let's find the partial derivatives:

$$f_x^{\prime }(x,y)=(x^2{e}^{2y}-y^2{e}^{2x}{)}_x^{\prime }=2x{e}^{2y}-2y^2{e}^{2x};$$

$$f_y^{\prime }(x,y)=(x^2{e}^{2y}-y^2{e}^{2x}{)}_y^{\prime }=2x^2{e}^{2y}-2y{e}^{2x}.$$

From here, we find

$$\frac{dy}{dx}=-\frac{f_x^{\prime }(x,y)}{f_y^{\prime }(x,y)}=-\frac{2x{e}^{2y}-2y^2{e}^{2x}}{2x^2{e}^{2y}-2y{e}^{2x}}=\frac{y^2{e}^{2x}-x{e}^{2y}}{x^2{e}^{2y}-y{e}^{2x}}.$$

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

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

Solution.

We find the derivative $\frac{dy}{dx}$ using the formula$$\frac{dy}{dx}=-\frac{f_x^{\prime }(x,y)}{f_y^{\prime }(x,y)}.$$ Here $f(x,y)=x-y+arctgy.$

Let's find the partial derivatives:

$$f_x^{\prime }(x,y)=(x-y+arctgy{)}_x^{\prime }=1;$$

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

From here, we find

$$\frac{dy}{dx}=-\frac{f_x^{\prime }(x,y)}{f_y^{\prime }(x,y)}=-\frac{1}{-1+\frac{1}{1+y^2}}=\frac{1+y^2}{y^2}.$$

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

$$\frac{d^{2}y}{d{x}^2}=\frac{(1+y^2{)}_x^{\prime }{y}^2-(y^2{)}_x^{\prime }(1+y^2)}{y^4}=\frac{2y{y}_x^{\prime }{y}^2-2y{y}_x^{\prime }(1+y^2)}{y^4}=-2\frac{y_x^{\prime }}{y^3}=$$ $$=-2\frac{\frac{1+y^2}{y^2}}{y^3}=-2\frac{1+y^2}{y^5}$$

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

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

Solution.

We find the derivatives $\frac{dz}{dx}$ and $\frac{dz}{dy}$ using the formulas$$\frac{dz}{dx}=-\frac{f_x^{\prime }(x,y,z)}{f_z^{\prime }(x,y,z)};$$ $$\frac{dz}{dy}=-\frac{f_y^{\prime }(x,y,z)}{f_z^{\prime }(x,y,z)};$$ Here $f(x,y,z)=x+y+z-e^z.$

Let's find the partial derivatives:

$$f_x^{\prime }(x,y,z)=(x+y+z-e^z{)}_x^{\prime }=1;$$

$$f_y^{\prime }(x,y,z)=(x+y+z-e^z{)}_y^{\prime }=1;$$

$$f_z^{\prime }(x,y,z)=(x+y+z-e^z{)}_z^{\prime }=1-e^z.$$

From here, we find

$$\frac{dz}{dx}=-\frac{f_x^{\prime }(x,y,z)}{f_z^{\prime }(x,y,z)}=-\frac{1}{1-e^z}.$$

$$\frac{dz}{dy}=-\frac{f_x^{\prime }(x,y,z)}{f_z^{\prime }(x,y,z)}=-\frac{1}{1-e^z}.$$

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

$$\frac{d^{2}z}{d{x}^2}=\frac{(1-e^z{)}_x^{\prime }}{(1-e^z{)}^2}=-\frac{e^z{z}_x^{\prime }}{(1-e^z{)}^2}=-\frac{-e^z\cdot \frac{1}{1-e^z}}{(1-e^z{)}^2}=\frac{e^z}{(1-e^z{)}^3}.$$ $$\frac{d^{2}z}{dxdy}=\frac{(1-e^z{)}_y^{\prime }}{(1-e^z{)}^2}=-\frac{e^z{z}_y^{\prime }}{(1-e^z{)}^2}=-\frac{-e^z\cdot \frac{1}{1-e^z}}{(1-e^z{)}^2}=\frac{e^z}{(1-e^z{)}^3}.$$ $$\frac{d^{2}z}{d{y}^2}=\frac{(1-e^z{)}_y^{\prime }}{(1-e^z{)}^2}=-\frac{e^z{z}_y^{\prime }}{(1-e^z{)}^2}=-\frac{-e^z\cdot \frac{1}{1-e^z}}{(1-e^z{)}^2}=\frac{e^z}{(1-e^z{)}^3}.$$

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

Tags: calculus, complex functions, derivative, functions of several independent variables, mathematical analysis