To find the derivative of a function in this form, we must make use of the chain rule, which states that for a give function: 

Its derivative is defined as:

In this case we have the function:

which is in the form of the composite function

in which

 ,     ,     

By similar logic, the derivative of the composite function is 

in which

and  

As a result, we have that 

The Product Rule of derivatives states that for a given function:

The derivative is defined as 

In this case, for the given function

 , 

and the respective derivatives are:

 , 

Applying the product rule we get that

To find the first derivative of this function, we must make use of the Quotient Rule of derivatives. That is, for a function

The derivative is defined as 

In this case 

 , 

 , 

As a result, 

First we note that 

.  

Now we take the derivative  of the function to find the slope of the tangent line.

Using the chain rule we get 

.  

Now we want to find the specific slope of the tangent line containing the .  

We then find that the slope is 

.  

We now plug in these values to slope-intercept form to find the y-intercept of the tangent line. 

which gives us 

.  

Thus the equation is simply 

.

To find the local max, you must find the first derivative, which is .

Then. you need to set that equal to zero, so that you can find the critical points. The critical points are telling you where the slope is zero, and also clues you in to where the function is changing direction. When you set this derivative equal to zero and factor the function, you get , giving you two critical points at  and .

Then, you set up a number line and test the regions in between those points. To the left of -1, pick a test value and plug it into the derivative. I chose -2 and got a negative value (you don't need the specific number, but rather, if it's negative or positive). In between -1 and 1, I chose 0 and got a possitive value. To the right of 1, I chose 2 and got a negative value. Then, I examine my number line to see where my function was going from positive to negative because that is what yields a maximum (think about a function going upwards and then changing direction downwards). That is happening at x=1.

When the derivative of a function is equal to zero, that means that the point is either a local maximum, local miniumum, or undefined. The derivative of  is . The derivative of the given function is 

We must now set it equal to zero and factor.

Now we must plug in points to the left and right of the critical points to determine which is the local maximum. 

This means the local maximum is at  because the function is increasing at numbers less than -2 and decreasing at number between -2 and 6

The points where the derivative of a function are equal to 0 are called critical points. Critical points are either local maxs, local mins, or do not exist. The derivative of  is . The derivative of the function is

Now we must set it equal to 0 and factor to solve.

We must now plug in points to the left and right of the critical points into the derivative function to figure out which is the local max.

This means that the function is increasing until it hits x=-6, then it decreases until x=1, then it begins increasing again.

This means that x=-6 is the local max.

At local maximums and minumims, the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative of the function.

The derivative of  is . Thus the derivative of the function is 

To find maximums and minumums we set it equal to 0.

So the critical points are at x=1 and x=2. To figure out the maximum we must plug each into the original function.

So the local max is at x=1.

By graphing the equation , we can see that there minimum at , and that the graph continues to rise in both directions around this point, so this must be a local minimum. We also know that the graph rises infinitely in both directions, so this must be the only local minimum.

Another way to identify the local minima is by taking the derivative of the function and setting it equal to zero.

Using the power rule,

 we find the derivative to be,

.

From here we set the derivative equal to zero and solve for x. By doing this we will identify the critical values of the function

Now we will plug in the x value and find the corresponding y value in the original equation. We will also plug in an x value that is lower than the critical x value and a x value that is higher than the critical value to confirm whether we have a local minima or maxima.

Since both of the x values have a larger y value than the y value that corresponds to , we know that the minimum occurs at .

The points where the derivative of the function is equal to 0 are called critical points. They are either local maximums, local minimums, or do not exist. The derivative of  is . The derivative of the function is

.

We must now set it equal to zero and factor to solve.

Now we must plug in points to the left and right of the critical points into the derivative function to find the local min.

This means the function is increasing until it hits x=2, then it decreases until it hits x=4 and begins increasing again. This makes x=4 the local minumum