All Calculus 1 Resources
Example Questions
Example Question #1 : Slope
What is the slope of the tangent line of f(x) = 3x4 – 5x3 – 4x at x = 40?
743,996
331,841
684,910
None of the other answers
768,000
743,996
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
Example Question #1 : Slope
Find the slope of the line tangent to when is equal to .
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
Example Question #1 : How To Find Slope By Graphing Functions
Find function which gives the slope of the line tangent to .
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.
Example Question #2 : How To Find Slope By Graphing Functions
Find the slope of the line tangent to at .
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.
Example Question #3 : How To Find Slope By Graphing Functions
Find the slope of at .
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
Example Question #3 : Slope
Find the slope of at given . Assume the integration constant is zero.
The first step here is to integrate in order to get .
Here the problem tells us that the integration constant , so
Plug in here
Example Question #3 : Slope
Consider the curve
.
What is the slope of this curve at ?
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 .
Example Question #2 : Slope
Find the line tangent to at .
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:
Example Question #9 : Slope
An isosceles triangle has one point at , one point at and one point on the -axis. What is the slope of the line between the point on the -axis and ?
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.
Example Question #5 : Slope
What is the slope of the line tangent to the graph of at ?
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 .