Slope - AP Calculus AB

The slope (m) between two points is found with the following formula:

We can apply this formula with the points we are given:

This is one of the answer choices.

To find the value of the slope at , you must first find the derivative of the function since that will give us the slope. To take the derivative of a term, multiplty the exponent by the coefficient in front of the term, and then subtract from the exponent. Therefore, the derivative is: . Then, plug in to get the correct slope value. .

To find the slope, you must first find the derivative. To take the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent. Therefore, the derivative is: . Then, plug in 2 to get the specific value: .

To find the slope, you must first find the derivative function. To take the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent. Therefore, the derivative is: . Then, plug in -1. .

To find the slope, you must first find the derivative of the function. To take the derivative, multiply the exponent by the coefficient in front of the x term and then subtract 1 from the exponent: . Now, plug in 1 for x to get your answer of x=5.

Find the slope of the line tangent to the curve of g(x) when x is equal to 5.

To find the slope of a tangent line, first find the derivative of the beginning functions:

Will become:

Next, simply plug in 5 everywhere we have an x and solve.

So our answer is 55548

Find the slope of the line tangent to h(x), when .

To find the slope of a tangent line, we need to find the derivative of our function:

Begin by recalling the rule for polynomial derivatives, derivative of sine, and derivative of

Polynomials derivatives are found by decreasing our exponent by one, and then dividing by that number.

The derivative of sine is cosine

The derivative of is

So with that in mind, let's find h'(x)

Next, we need to find h'(0), so plug in 0 for x and simplify:

So our slope is 0 when x=0.

Find the slope of the line tangent to f(x) when x is -5.

To find the slope of a tangent line, first find the derivative. Then, plug in the given point. Recall the power rule of derivatives. (Multiply each term by its exponent, then subtract one from the exponent)

Becomes:

Next, plug in -5

In doing so, we arrive at -112530, a very steep slope indeed!

The first derivative is easy:

f'(x) = 12x3 – 15x2 – 4

The slope of the tangent line is found by calculating f'(40) = 12 * 403 – 15 * 402 – 4 = 768,000 – 24,000 – 4 = 743,996

To find the slope of a tangent line, we need to find the first derivative of the function at that point. In other words, we need y'(6).

Taking the first derivative using the Power Rule we get the following.

Substituting in 6 for b and solving we get:

.

So our answer is 320160

To find the slope of a tangent line, we need the first derivative.

Recall that to find the first derivative of a polynomial, we need to decrease each exponent by one and multiply by the original number.

The slope of the tangent line can be found easily via derivatives. To find the slope of the tangent line at s=16, find b'(16) using the power rule on each term which states:

Applying this rule we get:

Therefore, the slope we are looking for is 454.

To find the slope of the line at that point, find the derivative of f(x) and plug in that point.

Remember that the derivative of and the derivative of

Now plug in

The first step here is to integrate in order to get .

Here the problem tells us that the integration constant , so

Plug in here

The slope of a curve at any point is equal to the derivative of the curve at that point.

Remembering that the derivative of and using the power rule on the second term we find the derivative to be:

.

Pluggin in we find that the slope is .

The other point of the triangle must be at as it must be equidistant from the other two points of the triangle. Since all points on the y-axis are units away from the other points in the direction, the third point must be equidistant in the direction from both and . The distance between these points is , so the third point must have a y-value of . The third point is now at so the slope of the line from to is as follows.

We must take the derivative of the function using the chain rule yielding .

The chain rule is .

Also remember that the derivative of is .

Applying these rules we get the following.

Plugging in the value for we get which is .

Find the line tangent to at .

First, we find :

Next, we find the derivative:

Therefore, the slope at is:

.

Using point-slope form, we can write the tangent line:

Simplifying this gives us:

A line is tangent to a point on a curve when its slope is equivalent to the slope of the curve at the point of intersection. Therefore to solve this equation, the slope of the curve at must be found.

The slope of a graph at any point can be found by taking the first derivative. To take the derivative of this equation, we must use the power rule,

.

We also must remember that the derivative of an constant is 0. By taking the first derivative of the graph equation, we obtain the slope equation

.

Plugging in , we find the slope at that point is 13, therefore any line tangent to the curve at that point must have a slope of 13.

To find the a general formula for the slope of the function , derive the function with respect to :

The slope of the function at is then found as: