This problem is asking you to find the indefinite integral of the given function, f'(x). To do this, we need to work through the function one term at a time. For each term in the function, we perform two steps. First, we divide the coefficient of that term by (n+1), where "n" is the value of the exponent for that term's x-value. Second, we add 1 to the value of that term's x-value. We continue this process for each term in the function. 

So, for , this would look like

which when integrated will equal 

You must remember to include "C" in your final answer; this acts as a placeholder for any constants that may have been present in the original equation.

This problem is asking you to find the indefinite integral of the given function, f'(x). To do this, we need to work through the function one term at a time. For each term in the function, we perform two steps. First, we divide the coefficient of that term by (n+1), where "n" is the value of the exponent for that term's x-value. Second, we add 1 to the value of that term's x-value. We continue this process for each term in the function. 

So, for , this would look like

which when integrated will equal 

You must remember to include "C" in your final answer; this acts as a placeholder for any constants that may have been present in the original equation.

We know that since its a half life question, the exponential used will be:

 where  relates to length of the half life. Since the half life is 3 days in this question, . 

Since we know how the mass is decaying over time, we simply need to put our  and  properly such that:

.

To check, if you plug , we get that , which indicates that we've lost half the mass in 3 days. 

To determine this, we take the derivative of : 

If we set , where  is a function that sets the two sides equal:

Notce that if we multiply the right side by , and set , we get that:

Backtracking we get that:

Although this may seem difficult, since both sides have identical integrals, we can just compare what's underneath the integral. 

Moving to the right:

We know that:

, since acceleration is the 2nd derivative of position. Knowing this we plug this into the formula to get that: 

The equation of the line tangent to the curve of f(x) at x=2 will have a slope equal to the instantaneous rate of change of the curve at x=2. In other words, the slope of the tangent line will have a slope equal to the derivative of f(x) defined at x=2, namely f'(2).

To solve this, we have that

  ;  

  ;  

To get the equation of the tangent line, we must use the Point-Slope formula utilizing the slope obtained from the derivative and the point of tangency.

The Point-Slope formula is defined as

in which m is the slope obtained from the derivative, m=23

and (x1,y1) is the point of tangency, (2,34). We then have that

Converting it to slope-intercept form, we have that

The equation of the normal line is defined as the line that is perpendicular to the line tangent to the curve at the point of tangency, which occurs at x=2 in this case.

The line perpendicular to the tangent line will have a slope equal to the negative reciprocal of the slope of the tangent line.

If the slope of the tangent line is denoted by the variable n, the normal line will have a slope (m) in which

We begin by defining the point of tangency (x1,y1) and the slope of the tangent line, n. 

  ; 

  ; 

We then have that the slope of the normal line, m is

Utilizing the Point-Slope Formula y-y1=m(x-x1), we have that

Converting to Slope-Intercept form, we get that the slope of the normal line is

To evaluate higher order derivative of trigonometric functions like the tangent function, we must become familiar with all of the other derivation rules; in this case we will use the Chain Rule and the Product Rule. 

We begin with the identity that 

Since the first derivative has now turned into a composite function in the form

 where    and   

and      and  

we have that the second derivative, utilizing the Chain Rule, is

To obtain the third derivative, we must derive the second derivative via the Product Rule, since there is the multiplication of u=2[sec(x)]^2 and v=tan(x).

Recall that the Product Rule states that for a function, f(x)=uv:

As a result we have that

In order to solve for the derivative of an equation that is not written as a function of a single variable (x), but rather two variables (x and y), we must implement Implicit Differentiation. 

Implicit Differentiation states that when the derivative of a function in terms of one variable (in this case, y) is taken with respect to the other variable present (in this case,x), one must multiply that derived term, f'(y) by the by the derivative (y' or dy/dx). 

In other words, if we have a function in terms of y, we have that it's derivative with respect to x is 

In this case, we must derive both sides of the equation, so we begin with the left side f(x,y)=6xy. Since two terms are being multiplied, we must implement the product rule. We then have that

Then we look at the right side of the equation f(x,y)=x^2+y^2 and we have that

Putting together the left and the right side of the equations, we have that

Algebraically, we simplify this equation to

To isolate our derivative term (dy/dx), we divide the left term by (2y-6x) to get: