The local maximums and minumums of a function are where the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative. Then we must set ot equal to 0 and solve. The derivative of  is . 

The critical points are at the above two points. To find the minimum we must plug both back into the origianl function.

Thus the local min is at x=-2.

To find the minimum of a function, start by finding the critical points of that function, or points where the derivative is equal to zero. Use the power rule to find the derivative:

Applying the power rule to the given equation, noting the constants in the first and second terms:

Then check to see if the critical point is a maximum, minimum, or an inflection point by taking the second derivative, using the power rule once again.

Because the second derivative is positive, the critical point  is a minimum.

To find the point where the minimum occurs, plug  back into the original equation and solve for .

Therefore, the minimum is 

The local minimum of a function can be found by finding the derivative and graphing it. The point in which the x axis is crossed from below gives the x position where the local minimum is found. Taking the derivative:

The graph of the derivative is shown below:

As shown by the graph, the local minimum is found at x = -4.

In order to determine the graph by inspection, there are key features to look for. The most important is the locations of the local maxima and minima in the graph of f(x). These points correspond to the x-intercepts in the graph of the derivative. Taking a look at the graph of f(x), you can see that the x intercepts on the graph of f'(x) will be located roughly at x = -3 and x = 4.5. Looking at the possible answers, the only two that could be graphs of f'(x) are these two:

 and 

The next step would then be to see which corresponds correctly to maxima and minima. Since the point at x = -3 is a local maximum, f(x) will increase up until the point at which it is maximum, then begin to drop. As seen in the positively oriented parabola, the rate of change of f(x) (the derivative) is positive up until it reaches x = -3. This means that f(x) was increasing, and indicates that this point was a local maximum. On the other hand, if you look at the graph on the left with the negatively oriented parabola, f'(x) is negative until it reaches the local maximum, which doesn't make sense, since that would mean it was decreasing up until the point and then increasing. This indicates a minima.

Since the point at x = -3 is a local maximum, the only graph that could be the derivative of f(x) is the positively oriented parabola. 

We'll solve this using the chain rule.

D_x[(5+3x)⁵]

=5(5+3x)⁴ * D_x[5+3x]

=5(5+3x)⁴(3)

=15(5+3x)⁴

First, remember that D_x[sin(x)]=cos(x). Now we can solve the problem using the Chain Rule.

D_x[sin(7x)]

=cos(7x)*D_x[7x]

=cos(7x)*(7)

=7cos(7x)

We can calculate this answer in steps.  We start with differentiating in terms of the left most variable in "xxyz".  So here we start by taking the derivative with respect to x. 

First, f_x= 4cos(4x+yz)

Then, f_xx= -16sin(4x+yz)

f_xxy= -16zcos(4x+yz)

Finally, f_xxyz= -16cos(4x+yz) + 16yzsin(4x+yz) 

 thus: 

thus:

We can use separation of variables to solve this problem since all of the "y-terms" are on one side and all of the "x-terms" are on the other side.  The equation can be written as .  

Integrating both sides gives us .