All Calculus 1 Resources
Example Questions
Example Question #242 : Graphing Functions
Find the equation for the line tangent to the curve at .
The derivative of the function is , and is found using the power rule
and the rule for the derivative of natural log which is,
so plugging in gives , which must be the slope of the line since the tangent line's slope is determined by the derivative.
Thus, the line is of the form , where b is unknown.
Solve for b by setting the equation equal to and plugging in for x since that is the given point.
, which gives us
Example Question #11 : How To Find Equation Of Line By Graphing Functions
Find the slope of the tangent line through the given point of the following function.
at the point
In order to find the slope of the tangent line through a certain point, we must find the rate of change (derivative) of the function. The derivative of is written as . This tells us what the slope of the tangent line is through any point in our function . In other words, all we need to do is plug-in (because our point has an x-value of 1) into . This will give us our answer, .
Example Question #17 : Equation Of Line
Find the tangent line to the function at the point .
To find the tangent line one must first find the slope, this can be given by the derivative evaluated at a point.
To find the derivative of this function use the power rule which states,
The derviative of is .
Evaluated at our point ,
we find that the slope, m is also 3.
Now we may use the point-slope equation of a line to find the tangent line.
The point slope equation is
Where is the point at which the line is tangent.
Using this definition we find the tangent line to be defined by .
Example Question #1 : Slope
What is the slope of the tangent line of f(x) = 3x4 – 5x3 – 4x at x = 40?
None of the other answers
768,000
331,841
684,910
743,996
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 #2721 : Calculus
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 #2722 : Calculus
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 : Slope
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 #5 : How To Find Slope By Graphing Functions
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 #5 : How To Find Slope By Graphing Functions
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 .