First we simplify the function using properties of logarithmic functions:

\(\displaystyle ln(x^a)=aln(x)\)   and  \(\displaystyle ln(\frac{x}{y})=ln(x)-ln(y)\)

Therefore:

\(\displaystyle y=ln({\sqrt[3]x{}})+ln(\frac{1}{x})=\frac{1}{3}ln(x)+ln(1)-ln(x)\)

also

\(\displaystyle ln(1)=0\)

Therefore    

\(\displaystyle y=\frac{1}{3}ln(x)-ln(x)\)

and

\(\displaystyle y'=\frac{1}{3}\frac{1}{x}-\frac{1}{x}=\frac{1-3}{3x}=-\frac{2}{3x}\)

to derive this equation we use the product rule:

\(\displaystyle (f\cdot g)'= f'\cdot g+g'\cdot f\)

\(\displaystyle f=cos(x)\)   \(\displaystyle f'=-sin(x)\)

and

\(\displaystyle g=e^{5x}\)  \(\displaystyle g=5e^{5x}\)

Therefore:

\(\displaystyle (cos(x)\cdot e^{5x})'=-sin(x\cdot e^{5x})+5cos(x)\cdot e^{5x}\)

\(\displaystyle =e^{5x}(5cos(x)-sin(x))\)

\(\displaystyle \small \small y= \frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2}\)

Logarithmic differentiation exploits the properties of logarithms to easily compute derivatives for functions that would otherwise be extremely tedious to find. Direct differentiation using the quotient rule could become quite messy. Take the natural logarithm of both sides of the equation, 

\(\displaystyle \small \ln (y)= \ln\left(\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2}\right)\)                               (1)

Expand the right-side using the properties of logarithms:

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Properties of Logarithmic Functions:

1. \(\displaystyle \small \ln(ab)=\ln(a)+\ln(b)\)

2. \(\displaystyle \small \ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)\)

3. \(\displaystyle \small \ln(a^n)=n\ln(a)\)

Then proceed with the differentiation using the known derivative of the natural logarithm function and the chain rule: 

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Derivative of the Natural Logarithm 

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

For a function \(\displaystyle \small y\) of \(\displaystyle \small x\), apply the chain-rule, 

\(\displaystyle \small \frac{d}{dx}\ln(y)-\frac{y'}{y}\)

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Expanding the right-side of equation (1) first by using Property 2. 

\(\displaystyle \small \ln(y)=\ln\left(\sqrt{2x^2-1}\right )-\ln\left[(x^3+1)(2x^2+1)^2 \right ]\)

Expand the second term with Property 1. Use Property 3 to pull out the exponent in the third term obtained after applying Property 1. 

 \(\displaystyle \small \small \small \small \ln(y)=\ln\left(\sqrt{2x^2-1}\right )-\ln\left(x^3+1)-2\ln\left[\left(2x^2+1)\right\right]\)

Differentiating implicitly over both sides of the equation with respect to \(\displaystyle \small x\). Be sure to apply the chain rule as needed. 

\(\displaystyle \small \small \small \frac{y'}{y}=\frac{1}{\sqrt{2x^2-1}}(4x)-\frac{1}{x^3+1}(3x^2+1)-\frac{2}{2x^2+1}(4x)\)

So now the derivative we were looking for, \(\displaystyle \small y'\) can be solved by multiplying both sides by \(\displaystyle \small y\) and then substituting back in the original function to write everything in terms of \(\displaystyle \small x.\)

 \(\displaystyle \small \small \small y'=\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2} \left(\frac{4x}{\sqrt{2x^2-1}}-\frac{3x^2+1}{x^3+1}-\frac{8x}{2x^2+1} \right )\)

The farmer's fencing material needs to cover the perimeter of his property. Since this piece of property is shaped like a rectangle, we know that the perimeter can be modeled with the equation

\(\displaystyle 2l+2w=P\).

In this case, we know that \(\displaystyle P=700\), since the \(\displaystyle 700\) ft. of fencing need to fit around the whole property.

This problem wants to maximize the area, so we're trying to find which values maximize this equation:

\(\displaystyle A=lw\).

We know that 

\(\displaystyle 2L+2w=700\), or simplified, that \(\displaystyle l+w=350\).

Solving for \(\displaystyle w\) gives us 

\(\displaystyle w=350-l\),

which we can plug into our area equation, giving us

\(\displaystyle A=l(350-l)\) 

\(\displaystyle A=350l-l^{2}\).

Taking the first derivative gives us 

\(\displaystyle A'=350-2l\).

Making \(\displaystyle A'\) equal zero allows us to solve for \(\displaystyle l\). 

\(\displaystyle 0=350-2l\).

So, \(\displaystyle l\) is 175 ft. To determine if this value is a maximum length, or a minimum, we take the second derivative of our area equation, which yields a constant\(\displaystyle -2\). Because this value is always less than zero, 175 ft. is a maximum. using our perimeter formula, we see that \(\displaystyle w\) is also equal to 175 ft. So, the maximum area the farmer can fence off is 175 ft. x 175 ft., or 30,625  sq. ft.

on this problem we apply the product rule:

let:  \(\displaystyle f=x^3\)   \(\displaystyle f{}'=3x^2\)  and  \(\displaystyle g=e^x\)   \(\displaystyle g'=e^x\)

\(\displaystyle \therefore y'=(x^3 \cdot e^x)'= 3x^2\cdot e^x + e^x\cdot x^3\)

\(\displaystyle =3x^2e^x+e^xx^3=x^2e^x(x+3)\)

to find the derivative of this function we need to use the chain rule:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\)   

let

\(\displaystyle u=e^{3x}\)   and   \(\displaystyle y=sin(u)\)   \(\displaystyle \therefore \frac{dy}{du}=cos(u)=cos(e^{3x})\)  

and 

\(\displaystyle \frac{du}{dx}=3e^{3x}\)

and

\(\displaystyle \frac{dy}{du}=3e^{3x}cos(e^3x)\)\(\displaystyle \frac{dy}{dx}=3e^{3x}cos(e^{3x})\)

We start at x = 0 and move to x=2, with a step size of 1. Essentially, we approximate the next step by using the formula:

\(\displaystyle p(x + \Delta x) = p(x) + p'(x)\cdot (\Delta x)\).

So applying Euler's method, we evaluate using derivative: 

\(\displaystyle p'(x) = 0.5x(1-x), \ p(0) = 2\) 

And two step sizes, at x = 1 and x = 2.

\(\displaystyle {}p(x=1) = p(0) + (1 - 0)p'(0) = 2 + 1\cdot p'(0) = 2 + 1 \cdot 0.5(0)(1-0) = 2 + 0 = 2\) 

\(\displaystyle {}p(x=2) = p(1) + (2-1)\cdot p'(1) = 2 + 1\cdot p'(1) = 2 + 0.5(1)(1-1) = 2 + 0 = 2\)

And thus the evaluation of p at x = 2, using Euler's method, gives us p(2) = 2.

Using Euler's method with \(\displaystyle \small h=1/4\) means that we use two iterations to get the approximation. The general iterative formula is 

\(\displaystyle \small y_{i+1}=y_i+h\cdot f(t_i,y_i)\)

where each \(\displaystyle \small t_i\) is

\(\displaystyle \small t_0=0\)

\(\displaystyle \small t_1=\frac{1}{4}\)

\(\displaystyle \small t_2=\frac{1}{2}\)

 \(\displaystyle \small y_i\) is an approximation of \(\displaystyle \small y(t_i)\), and \(\displaystyle y'=f(t,y)=y\), for this differential equation. So we have

\(\displaystyle \small \small \small \small y_1=y_0+\frac{1}{4}f(t_0,y_0)=1+\frac{1}{4}\cdot 1=1.25\)

\(\displaystyle \small \small \small \small e^{1/2}= y(\frac{1}{2})\approx y_2=y_1+\frac{1}{4}f(t_1,y_1)=1.25+\frac{1}{4}\cdot 1.25=1.5625\)

So our approximation of \(\displaystyle \small e^{1/2}\) is

\(\displaystyle \small e^{1/2}\approx 1.5625\)

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

\(\displaystyle x_{n+1}=x_n+h\)

\(\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)\)

are solved starting at the initial condition and ending at the desired value.  \(\displaystyle f(x_n,y_n)\) is the solution to the differential equation.

In this problem,

\(\displaystyle f(x,y)=\frac{dy}{dx}=3x+4y\) 

\(\displaystyle h=0.25\)

Starting at the initial point \(\displaystyle (x_0,y_0)=(0,0)\)

\(\displaystyle x_{1}=x_0+h=0+0.25=0.25\)

\(\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0+0.25*(3*0+4*0)=0\)

\(\displaystyle (x_1,y_1)=(0.25,0)\)

\(\displaystyle x_{2}=x_1+h=0.25+0.25=0.5\)

\(\displaystyle y_{2}=y_1+h*f(x_1,y_1)=0+0.25*(3*0.25+4*0)=0.1875\)

\(\displaystyle (x_2,y_2)=(0.5,0.1875)\)

We continue using Euler's method until \(\displaystyle x=1\).  The results of Euler's method are in the table below.

Note: Solving this differential equation analytically gives a different answer, \(\displaystyle 9.2997\).  Future problems will explain this discrepancy. 

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

\(\displaystyle x_{n+1}=x_n+h\)

\(\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)\)

are solved starting at the initial condition and ending at the desired value.  \(\displaystyle f(x_n,y_n)\) is the solution to the differential equation.

In this problem,

\(\displaystyle f(x,y)=\frac{dy}{dx}=3x+4y\) 

\(\displaystyle h=0.1\)

Starting at the initial point \(\displaystyle (x_0,y_0)=(0,0)\)

\(\displaystyle x_{1}=x_0+h=0+0.1=0.1\)

\(\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0+0.1*(3*0+4*0)=0\)

\(\displaystyle (x_1,y_1)=(0.1,0)\)

\(\displaystyle x_{2}=x_1+h=0.1+0.1=0.2\)

\(\displaystyle y_{2}=y_1+h*f(x_1,y_1)=0+0.1*(3*0.1+4*0)=0.03\)

\(\displaystyle (x_2,y_2)=(0.2,0.03)\)

We continue using Euler's method until \(\displaystyle x=1\).  The results of Euler's method are in the table below.

Note: Solving this differential equation analytically gives a different answer, \(\displaystyle 9.2997\). As the step size gets larger, Euler's method gives a more accurate answer.