Numerical Methods

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 1.4 – The Approximation Method of Euler

Expected Educational Results

• Objective 1–1: I can evaluate or estimate a limit from the graph of a function.

• Objective 1–2: I can explain why a limit does not exist.

• Objective 1–3: I can explain the relationship between a limit that does not exist and the vertical asymptote of a function.

Prerequisites

Algebra

Equation of a Line

The point-slope of the equation of a line with slope $m$$m$ and passing through the point $(x_0,y_0)$$(x_0,y_0)$ is $y-y_0=m(x-x_0)$$y-y_0=m(x-x_0)$

Calculus I

Derivatives

Differentiation Rules

General Formulas

• ${\displaystyle \frac{d}{dx}c=0}$$\displaystyle \frac{d}{dx}c=0$, where $c$$c$ is any constant.

• ${\displaystyle \frac{d}{dx}[cf(x)]=c{f}^{\prime }(x)}$$\displaystyle \frac{d}{dx}[cf(x)]=cf^{\,\prime}(x)$, where $c$$c$ is any constant.

• ${\displaystyle \frac{d}{dx}[f(x)\pm g(x)]=f^{\prime }(x)\pm g^{\prime }(x)}$$\displaystyle \frac{d}{dx}[f(x)\pm g(x)]=f^{\,\prime}(x)\pm g^{\,\prime}(x)$

• ${\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+g^{\prime }(x)f(x)}$$\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\,\prime}(x)g(x)+g^{\,\prime}(x)f(x)$

• ${\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f^{\prime }(x)g(x)-g^{\prime }(x)f(x)}{[g(x){]}^2}}$$\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f^{\,\prime}(x)g(x)-g^{\,\prime}(x)f(x)}{[g(x)]^2}$

• ${\displaystyle \frac{d}{dx}f(g(x))=f^{\prime }(g(x))g^{\prime }(x)}$$\displaystyle \frac{d}{dx}f(g(x))=f^{\,\prime}(g(x))g^{\,\prime}(x)$

• ${\displaystyle \frac{d}{dx}{x}^n=n{x}^{n-1}}$$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$\index{differentiation rules!power rule}

Exponential and Logarithmic Formulas

• ${\displaystyle \frac{d}{dx}{e}^x=e^x}$$\displaystyle \frac{d}{dx}e^x=e^x$

• ${\displaystyle \frac{d}{dx}{a}^x=\ln (a)a^x}$$\displaystyle \frac{d}{dx}a^x=\ln{(a)}a^x$, $a>0$$a>0$, $a\ne 1$$a\ne 1$

• ${\displaystyle \frac{d}{dx}\ln (x)=\frac{1}{x}}$$\displaystyle \frac{d}{dx}\ln{(x)}=\frac{1}{x}$

• ${\displaystyle \frac{d}{dx}{\log }_b(x)=\frac{1}{x\ln (b)}}$$\displaystyle \frac{d}{dx}\log_b{(x)}=\frac{1}{x\ln{(b)}}$, $b>0$$b>0$, $b\ne 1$$b\ne 1$

Trigonometric Formulas

• ${\displaystyle \frac{d}{dx}\sin (x)=\cos (x)}$$\displaystyle \frac{d}{dx}\sin{(x)}=\cos{(x)}$

• ${\displaystyle \frac{d}{dx}\cos (x)=-\sin (x)}$$\displaystyle \frac{d}{dx}\cos{(x)}=-\sin{(x)}$

• ${\displaystyle \frac{d}{dx}\tan (x)={\sec }^2(x)}$$\displaystyle \frac{d}{dx}\tan{(x)}=\sec^2{(x)}$

• ${\displaystyle \frac{d}{dx}\cot (x)=-{\csc }^2(x)}$$\displaystyle \frac{d}{dx}\cot{(x)}=-\csc^2{(x)}$

• ${\displaystyle \frac{d}{dx}\sec (x)=\sec (x)\tan (x)}$$\displaystyle \frac{d}{dx}\sec{(x)}=\sec{(x)}\tan{(x)}$

• ${\displaystyle \frac{d}{dx}\csc (x)=-\csc (x)\cot (x)}$$\displaystyle \frac{d}{dx}\csc{(x)}=-\csc{(x)}\cot{(x)}$

Inverse Trigonometric Formulas

• ${\displaystyle \frac{d}{dx}\arcsin (x)=\frac{1}{\sqrt{1-x^2}}}$$\displaystyle \frac{d}{dx}\arcsin{(x)}=\frac{1}{\sqrt{1-x^2}}$

• ${\displaystyle \frac{d}{dx}\arccos (x)=-\frac{1}{\sqrt{1-x^2}}}$$\displaystyle \frac{d}{dx}\arccos{(x)}=-\frac{1}{\sqrt{1-x^2}}$

• ${\displaystyle \frac{d}{dx}\arctan (x)=\frac{1}{1+x^2}}$$\displaystyle \frac{d}{dx}\arctan{(x)}=\frac{1}{1+x^2}$

• ${\displaystyle \frac{d}{dx}\text{arccot}(x)=-\frac{1}{1+x^2}}$$\displaystyle \frac{d}{dx}\text{arccot}{(x)}=-\frac{1}{1+x^2}$

• ${\displaystyle \frac{d}{dx}\text{arcsec}(x)=\frac{1}{|x|\sqrt{x^2-1}}}$$\displaystyle \frac{d}{dx}\text{arcsec}{(x)}=\frac{1}{|x|\sqrt{x^2-1}}$

• ${\displaystyle \frac{d}{dx}\text{arccsc}(x)=-\frac{1}{|x|\sqrt{x^2-1}}}$$\displaystyle \frac{d}{dx}\text{arccsc}{(x)}=-\frac{1}{|x|\sqrt{x^2-1}}$

Hyperbolic Formulas

• ${\displaystyle \frac{d}{dx}\sinh (x)=\cosh (x)}$$\displaystyle \frac{d}{dx}\sinh{(x)}=\cosh{(x)}$

• ${\displaystyle \frac{d}{dx}\cosh (x)=\sinh (x)}$$\displaystyle \frac{d}{dx}\cosh{(x)}=\sinh{(x)}$

• ${\displaystyle \frac{d}{dx}\tanh (x)={\text{sech}}^2(x)}$$\displaystyle \frac{d}{dx}\tanh{(x)}=\text{sech}^2{(x)}$

• ${\displaystyle \frac{d}{dx}\coth (x)=-{\text{csch}}^2(x)}$$\displaystyle \frac{d}{dx}\coth{(x)}=-\text{csch}^2{(x)}$

• ${\displaystyle \frac{d}{dx}\text{sech}(x)=-\text{sech}(x)\tanh (x)}$$\displaystyle \frac{d}{dx}\text{sech}{(x)}=-\text{sech}{(x)}\tanh{(x)}$

• ${\displaystyle \frac{d}{dx}\text{csch}(x)=-\text{csch}(x)\coth (x)}$$\displaystyle \frac{d}{dx}\text{csch}{(x)}=-\text{csch}{(x)}\coth{(x)}$

Inverse Hyperbolic Formulas

• ${\displaystyle \frac{d}{dx}\text{arcsinh}(x)=\frac{1}{\sqrt{1+x^2}}}$$\displaystyle \frac{d}{dx}\text{arcsinh}{(x)}=\frac{1}{\sqrt{1+x^2}}$

• ${\displaystyle \frac{d}{dx}\text{arccosh}(x)=\frac{1}{\sqrt{x^2-1}}}$$\displaystyle \frac{d}{dx}\text{arccosh}{(x)}=\frac{1}{\sqrt{x^2-1}}$

• ${\displaystyle \frac{d}{dx}\text{arctanh}(x)=\frac{1}{1-x^2}}$$\displaystyle \frac{d}{dx}\text{arctanh}{(x)}=\frac{1}{1-x^2}$

• ${\displaystyle \frac{d}{dx}\text{arccoth}(x)=\frac{1}{1-x^2}}$$\displaystyle \frac{d}{dx}\text{arccoth}{(x)}=\frac{1}{1-x^2}$

• ${\displaystyle \frac{d}{dx}\text{arcsech}(x)=-\frac{1}{x\sqrt{1-x^2}}}$$\displaystyle \frac{d}{dx}\text{arcsech}{(x)}=-\frac{1}{x\sqrt{1-x^2}}$

• ${\displaystyle \frac{d}{dx}\text{arccsch}(x)=-\frac{1}{|x|\sqrt{1+x^2}}}$$\displaystyle \frac{d}{dx}\text{arccsch}{(x)}=-\frac{1}{|x|\sqrt{1+x^2}}$

Partial Derivatives

One of the conditions for the Existence and Uniqueness Theorem uses a first-order partial derivative of the dependent variable.

An ordinary differential equation (ODE) in the form ${\displaystyle \frac{dy}{dx}=f(x,y)}$$\displaystyle\frac{dy}{dx}=f(x,y)$ states that the derivative of $y$$y$ with respect to $x$$x$ is a function dependent on both $x$$x$ and $y$$y$, where $y$$y$ is a function of $x$$x$. Since the function $f$$f$ is dependent on more than one variable, we will need to find partial derivatives, specifically the partial derivative of $f$$f$ with respect to $y$$y$, i.e., ${\displaystyle \frac{\partial f}{\partial y}}$$\displaystyle\frac{\partial f}{\partial y}$ or $f_y(x,y)$$f_y(x,y)$. But this is the similar to derivatives (without needing implicit differentiation since $y$$y$ is a dependent variable of $f$$f$) in Calculus I if we treat all occurrences of $x$$x$ as either coefficients of $y$$y$ or constants.

Example 01

Find the partial derivative of $f(x,y)=x{y}^2-x^{3}y+x^2$$f(x,y)=x{\color{gsured}y^2}-x^3{\color{gsured}y}+x^2$ with respect to $y$${\color{gsured}y}$.

The solution is ${\displaystyle \frac{\partial f(x,y)}{\partial y}=x(2y)-x^3(1)+0=2xy-x^3}$$\displaystyle\frac{\partial f(x,y)}{\partial {\color{gsured}y}}=x({\color{gsured}2y})-x^3({\color{gsured}1})+0=2xy-x^3$.

Example 2

Find the partial derivative of $f(x,y)=x\sqrt[3]{y^2}-\tan (y)\sqrt{x}+15\ln (x)$$f(x,y)=x{\color{gsured}\sqrt[3]{y^2}}-{\color{gsured}\tan{(y)}}\sqrt{x}+15\ln{(x)}$ with respect to $y$${\color{gsured}y}$.

The solution is ${\displaystyle \frac{\partial f(x,y)}{\partial y}=x\left(\frac{2}{3}{y}^{-\frac{1}{3}}\right)-{\sec }^2(y)\sqrt{x}}$$\displaystyle\frac{\partial f(x,y)}{\partial {\color{gsured}y}}=x\left({\color{gsured}\frac{2}{3}y^{-\frac{1}{3}}}\right)-{\color{gsured}\sec^2{(y)}}\sqrt{x}$.

Practice Examples

Find the partial derivative of $f(x,y)$$f(x,y)$ with respect to $y$$y$. Verify your answers with a classmate or using software.

1. ${\displaystyle f(x,y)=x^2{y}^2-4x^5+{\sin }^2(x)-\ln (y)}$$\displaystyle f(x,y)=x^2y^2-4x^5+\sin^2{(x)}-\ln{(y)}$

2. ${\displaystyle f(x,y)=x\sin (y)-y\cos (x)}$$\displaystyle f(x,y)=x\sin{(y)}-y\cos{(x)}$

3. ${\displaystyle f(x,y)=\sqrt{x^2+y}}$$\displaystyle f(x,y)=\sqrt{x^2+y}$

4. ${\displaystyle f(x,y)=\frac{x}{y}}$$\displaystyle f(x,y)=\frac{x}{y}$

5. ${\displaystyle f(x,y)=\frac{y\sqrt{x}+4xy+e^{4y}}{x^2+1}}$$\displaystyle f(x,y)=\frac{y\sqrt{x}+4xy+e^{4y}}{x^2+1}$

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Last Modified: Monday, 31 August 2020 12:17 EDT