The first step is to find the derivative of the function given, which is .

Next, find the slope at  by plugging in  and solving for , which is the slope. You should get . This means the slope of the new line is also 3 because at the point where a slope and a line are tangent they have the same slope. Use the equation to express your line. Y and x are variables and m is the slope, so the only thing you need to find is b. Plug in the point and slope into  to get . Now you can express the general equation of the line as .

To find the equation of the line perpendicular to the tangent line to the function at a certain point, we must find the slope of the tangent line to the function, which is the derivative of the function at that point:

The derivative was found using the following rules:

Now, we evaluate the derivative at the given point:

We now know the slope of the tangent line, but because the line we are solving for is perpendicular to this line, its slope is the negative reciprocal, .

With a slope and a point, we can now find the equation of the line:

First, evaluate .

Then, to find the slope of the tangent line,  find .

, so .

Therefore, the equation of the tangent line is

.

In order to find the extreme value we need to take the derivative of the function.

After setting it equal to 0, we see that the only candidate is for . After setting  into , we get the coordinate  as an extreme value. To confirm it is a minimum we can plot the function.

To find the equation of a tangent line, we need two things: The tangent point, which is given as , and the slope of the tangent line at that point, which is the derivative at that point.

To find the derivative at the point, we will find , using derivative rules. Then we will plug the given point's x-coordinate into and that will give us the slope we need.

Finding the derivative of will require the power rule for each term. Recall that the power rule is . For the , the power rule effectively removes the . Also, the derivative of a constant is , so the will drop when we get the derivative.

Applying these rules, we get

Now that we have the derivative, we effectively have a formula to find the slope of a tangent line at any point we choose. The question asks for the tangent line at . So we plug the x-coordinate of the point into the derivative function to find the slope.

The last step is to use the point-slope equation of a line to construct the equation we need. The point-slope equation is , where is the slope, and is the given point.

Plugging our slope into , and our original point in for and , we get

Now we can solve for to compare to the answer choices.

This is the equation of the tangent line at the given point. We have our answer.

To find the equation of a tangent line at a point, we will need the slope of the function at that point. To find this, we find the derivative of the function.

Finding this derivative will use trigonometric function derivative rules, and the product rule.

Recall that the derivative of is . This takes care of the first term.

To find the derivative of the next term, we need to be careful with our signs. We will use the product rule which results in two terms. The negative sign can mess us up if we aren't careful. The way we will handle this is to associate the minus sign with the term when doing the product rule. So we will find the derivative of . The product rule is as follows, where the red and blue are the two factors of the term we are differentiating. In our case, and .

by the special case of the power rule. The drops.

by the trigonometric derivative rules.

Putting these together we get the derivative of to be

Simplifying, we get

Assembling all the pieces of the derivative together, we have found

Combining like terms gives

Now we have a formula for the tangent slope of at any point. The point we care about is . To find this point's tangent slope, we will plug its x-coordinate into our derivative.

Simplifying gives

So we have found the slope of the tangent line at our point, , is .

The last step is the point-slope equation of a line, , where is the slope and is the given point. Plugging in for , for , and for , we get

Solving for , we get

This is the equation of the tangent line at the given point. We have our answer.

Differentiate: 

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text. 

Let,

Therefore, 

Now we can write the derivative using the chain rule as: 

Let's calculate the derivative with respect to  in the second term using the Second FTOC and then convert back to . 

Therefore we have, 

Use the power rule to find the derivative.

To take the derivative you need to bring the power down to the front of the equation, multiplying it by the coefficient and then drop the power.

So:

  becomes 

because the degree of "x" is just one, and once you multiply 3 by 1 you get 3 and drop the power of "x" to 0. The second term is just a constant and the derivative of any term is just 0. 

To do this problem you need to use the quotient rule. So you do

(low)(derivative of the high)-(high)(derivative of the low) all divided by the bottom term squared.

So:

Which, when simplified is: