Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 1 and a rate of growth of 40:

\(\displaystyle \frac{dA}{dt}=12(1)(40)=480\)

To do rate of change \(\displaystyle m\), remember it is equivalent to finding slope. 

\(\displaystyle m=\frac{f(\pi)-f(0)}{\pi-0}=\frac{sec(2\pi)-sec(0)}{\pi-0} =\frac{1-1}{\pi-0} = 0\)

The average velocity of an object between \(\displaystyle t=a\) and \(\displaystyle t=b\) is given by the equation

\(\displaystyle v_{average}=\frac{\Delta y}{\Delta x}=\frac{r(b)-r(a)}{b-a}\)

In this problem,

\(\displaystyle r(4)=5*(4)^{2}-7(4)+6=80-28+6=58\)

\(\displaystyle r(2)=5*(2)^{2}-7(2)+6=20-14+6=0\)

\(\displaystyle v_{average}=\frac{r(4)-r(2)}{4-2}=\frac{58-0}{4-2}=29\)

The instantaneous velocity of the car is the first derivative of the position at a given point.

In this problem,

\(\displaystyle r(t)=5t^{2}-7t+6\)

\(\displaystyle v_{ins}=r'(t)=10t-7\)

\(\displaystyle v_{ins}=r'(4)=10*4-7=40-7=33\)

To solve this problem, we can use either the quotient rule or the product rule.  For this solution, we will use the product rule.  

The product rule states that \(\displaystyle \frac{d(u*v)}{dx}=u'v+v'u\).  

In this case, let \(\displaystyle u=x^{-8}\Rightarrow u'=-8x^{-9}\) and \(\displaystyle v=ln(x)\Rightarrow v'=\frac{1}{x}\).  

Putting both of these together, we get 

\(\displaystyle \frac{d(x^{-8}ln(x))}{dx}=-8x^{-9}ln(x)+\frac{1}{x}x^{-8}=\frac{1}{x^{9}}+\frac{-8ln(x)}{x^9}=\frac{1-8ln(x)}{x^9}\).

To find the slope of the tangent line at the specified point, we must first verify that the specified point actually exists on the curve. We check that 

\(\displaystyle 1=sin(x^2+y^2)=sin((\frac{\sqrt{\pi }}{2})^2+(\frac{\sqrt{\pi }}{2})^2)=sin(\frac{\pi }{2})=1\)

Since the verification checks out, the problem has a solution and we can continue with Implicit Differentiation. 

Recall that for Implicit differentiation, if we have a function in terms of y, we have that it's derivative with respect to x is 

\(\displaystyle \frac{d}{dx}[f(y)]=\frac{df}{dy}*\frac{dy}{dx}\)

Applying this to the given function, we have that 

\(\displaystyle \frac{d}{dx}[1=sin(x^2+y^2)]\)

We must also utilize the Chain Rule to obtain the derivative; we get that

\(\displaystyle 0=cos(x^2+y^2)*[2x+2y\frac{dy}{dx}]\)

Algebraically, we divide the cosine term to begin isolating dy/dx. We then get that

\(\displaystyle 0=2x+2y\frac{dy}{dx}\)

\(\displaystyle -2x=2y\frac{dy}{dx}\)

\(\displaystyle \frac{dy}{dx}=-\frac{x}{y}\)

To obtain the slope of the tangent line, we substitute the specified point (x,y) for x and y respectively. 

\(\displaystyle \frac{dy}{dx}=-\frac{\frac{\sqrt{\pi }}{2}}{\frac{\sqrt{\pi }}{2}}=-1\)

The volume of a sphere is 

\(\displaystyle V=\frac{4}{3}\pi r^3\) 

which can also be written as a function with respect to time, i.e.. 

\(\displaystyle V(t)=\frac{4}{3} \pi r(t)^3\).  

If we take the derivative of this, then 

\(\displaystyle V'(t)=4 \pi r(t)^2 \frac{dr}{dt}\).  

The problem tells us though that the rate of change of the radius is \(\displaystyle 5 cm/s\) and \(\displaystyle r=1 \text{cm}\).  

Plugging in these values we find that 

\(\displaystyle V'(t)=4 \pi (1cm)^2\times 5 cm/s= 20\pi cm^3/s\).

To do a linear approximation, we're going to create a function

\(\displaystyle y_1=mz+b\), that approximates our situation. In our case, m will be the slope of the function \(\displaystyle sec(\pi x)\) at \(\displaystyle x=2\), while b will be the value of the function \(\displaystyle sec(\pi x)\) at \(\displaystyle x=2\). The z will be distance from our starting position \(\displaystyle (x=2)\) to our end position \(\displaystyle (x=2.01)\), which is \(\displaystyle 0.01\). 

Firstly, we need to find the derivative of \(\displaystyle sec(\pi x)\) with respect to x to determine slope.

By the power rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} (cos(\pi x))^{-1}= -(cos(\pi x))^{-2}*-\pi sin(\pi x)=(cos(\pi x))^{-2}*\pi sin(\pi x)\)

The slope at \(\displaystyle x=2\) will therefore be 0 since \(\displaystyle sin(2\pi)=0\).

Since this is the case, the approximate value of our interest rate will be identical to the value of the original function at x=2, which is \(\displaystyle sec(2\pi)=1\). 

\(\displaystyle y_1=0*0.01+1=1\)

1 is our final answer. 

To do this, we must determine the slope of the function at \(\displaystyle x=0\), which we will call \(\displaystyle m\), and the initial value of the function at \(\displaystyle x=0\), which we will call \(\displaystyle b\), and since \(\displaystyle x=0.02\) is only \(\displaystyle .02\) away from \(\displaystyle x=0\), our linear approximation will look like:

\(\displaystyle y=m(.02)+b\)

To determine slope, we take the derivative of the function with respect to x and find its value at \(\displaystyle x=0\), which in our case is:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[3e^{-x}+sec(x)]= -3e^{-x}-sec(x)*tan(x).\)

At \(\displaystyle x=0\), our value for \(\displaystyle m\) is \(\displaystyle -3\)

To determine \(\displaystyle b\), we need to determine the value of the original equation at \(\displaystyle x=0\)

\(\displaystyle y=3e^{-x}+sec(x)\)

At \(\displaystyle x=0\), our value for b is \(\displaystyle 4\) 

Since \(\displaystyle y=m(.02)+b\), \(\displaystyle y=3.94\)

First recall that

\(\displaystyle ln(e^{-2})=-2\)

To find the tangent line of \(\displaystyle y=3e^{-x}\) at \(\displaystyle x=-2\), we first determine the slope of \(\displaystyle y=3e^{-x}\). To do so, we must find its derivative. 

Recall that derivatives of exponential functions involving \(\displaystyle e\) are given as:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}Ae^{g(x)}= A*g'(x)e^{g(x)}\), where \(\displaystyle A\) is a constant and \(\displaystyle g(x)\) is any function of \(\displaystyle x\)

In our case, \(\displaystyle A=3, g(x)=-x, g'(x)=-1\),. 

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

At \(\displaystyle x=-2\),

 \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=-3e^{-(-2)}}= -3e^2=m\), where \(\displaystyle m\) is the slope of the tangent line.

To use point-slope form, we need to know the value of the original function at \(\displaystyle x=-2\), 

\(\displaystyle y=3e^{-x}\)

\(\displaystyle y(-2)=3e^{-(-2)}=3e^2\)

Therefore,

\(\displaystyle (y_{appx}-3e^2)=-3e^2(x+2)\)

At \(\displaystyle x=ln(e^3)=3\), 

\(\displaystyle (y_{appx}-3e^2)=-3e^2(3+2)\)

\(\displaystyle y_{appx}=-15e^2+3e^2=-12e^2\)