The tangent line to the graph of  at point  is the line with slope  that passes through that point. Find the derivative :

Apply the sum rule:

The tangent line is therefore the line with slope 5 through . Apply the point-slope formula:

The tangent line to the graph of  at point  is the line with slope  that passes through that point. Find the derivative :

Apply the constant multiple and sum rules:

Set  and  and apply the chain rule. 

Substituting back:

Evaluate  using substitution:

The tangent line is therefore the line with slope  through .  is a -intercept, so apply the slope-intercept formula to get the equation

.

This is not among the choices given.

We first start by finding the slope of the line in question, which we do by taking the derivative of  and evaluate at . 

, 

We then use point slope form to get the equation of the line at the point 

The first step is to find the derivative of the function given, which is . Next, find the slope at (1,4) by plugging in x=1 and solving for , which is the slope. You should get . This means the slope of the new line is also -1 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 .

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 an 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 an answer.