Aline passes through the point (0, 1) and has a positive
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Answer:
The tangent line to the given curve at the given point is .
Step-by-step explanation:
To find the slope of the tangent line we to compute the derivative of and then evaluate it for
.
Differentiate the equation.
Differentiate both sides.
Sum/Difference rule applied:
Constant multiple rule applied:
Applied power rule:
Simplifying and apply constant rule:
Simplify.
Evaluate y' for x=4:
is the slope of the tangent line.
Point slope form of a line is:
where is the slope and
is a point on the line.
Insert 9 for and (4,10) for
:
The intended form is which means we are going need to distribute and solve for
.
Distribute:
Add 10 on both sides:
The tangent line to the given curve at the given point is .
------------Formal Definition of Derivative----------------
The following limit will give us the derivative of the function at
(the slope of the tangent line at
):
We are given f(4)=10.
Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:
Let's check this with FOIL:
First:
Outer:
Inner:
Last:
---------------------------------Add!
So the numerator and the denominator do contain a common factor.
This means we have this so far in the simplifying of the above limit:
Now we get to replace x with 4 since we have no division by 0 to worry about:
2(4)+1=8+1=9.