Card 0 of 20
What is the slope of the tangent line of f(x) = 3x4 – 5x3 – 4x at x = 40?
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
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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
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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.
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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.
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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
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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
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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
.
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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.
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Find the slope of the equation at the point
.
The slope of a function at a given point is found by first taking the derivative of the function.
Use the power rule to find this derivative, given by:
By evaluating the derivative when you will find the slope of the function at the point
.
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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
.
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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:
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What is the slope of a line tangent to at
?
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.
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What is the slope of the curve at the point
?
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:
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What is the slope of the tangent line to at
The derivative of a function describes the slope. Therefore, you must first find the derivative of the function, which is .
Then, you plug in the specific x value given in the problem, which is -1:
.
Therefore, your final answer is -7.
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At what values is the slope of the tangent line to
equal to zero?
First, you must find the slope equation of the tangent line to the function, which is just the derivative of the function:
.
Since that is the slope equation, you need to set that equal to 0 and factor:
, or
, which yields
.
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What is the slope of the line tangent to the function at ?
The slope of a tangent line to a function at a point can be found by taking the derivative of the function and plugging in the point at which the slope is to be found. The derivative of the funtion can be found using the product rule.
The derivative of is
.
Also, the derivative of a sin function is the cos function.
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Find the slope of the line tangent to the function at .
The slope of the tangent at a certain x is the value of the derivative at that point. The derivative of is
.
The derivative of the given function is
.
Plugging in x=3 gives
.
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If the equation for a graph given is , find the slope of a line tangent to this graph at
.
In order to find the slope, you take derivative of the graph equation. Then by plugging in any value, you can find the slope of the graph.
In order to take the derivative of equation, the power rule must be applied, . You must also apply the quotient rule
.
Taking the derivative of the graph equation
937,500+125-160
Plugging in , you find the slope to be
.
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The coordinates of the following points are given as follows: ,
and
.
If a is tangent to
at point
and
is tangent to
at point
. If
, then which of the following statements is true?
If the two lines are perpendicular, then their slopes must be opposite reciprocals. Since both lines are tangent to the curve, , their slopes will be equal to the slope of the curve at the points to which they are tangent. So
has slope
and
has slope
. Following this, if the lines are perpendicular, then
.
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Find the slope of the following function at .
This problem basically amounts to finding the derivative and evaluating it at the given value. We need to use a couple different techniques to find the derivative of this function but they're all fairly simple. We need the chain rule which says:
and the product rule, which is
So, using these we can calculate our derivative. and
, so:
this equals,
and when we plug in , we get
which can be written as .
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