How do you find the equation of a tangent line? how does it differ from a regular derivative. Give an example of finding an equation of a tangent line and an example of finding a regular derivative. Include the directions that would be given with each problem.
How do you find the equation of a tangent line?
ReplyDeleteTo find the equation of a tangent line you must take the derivative using the limit process which is when you plug the given equation into lim x-> 0 F(x + delta x) – F(x)/ delta x and then you must plug your results and the given points into the equation of a line formula which is y-y1 = m(x-x1) and once you solve for that formula you will have your equation of a tangent line.
how does it differ from a regular derivative?
A regular derivative would be solved by plugging the equation into the limit process which again, is when you plug the given equation into lim x-> 0 F(x + delta x) – F(x)/ delta x however you would stop working here and your answer would be the derivative, instead of plugging your results into formula for finding the equation of a tangent line.
In a nut shell, the major difference between finding the equation of a tangent line and finding a regular derivative is the format your answers must be resolved to.
The equation of a tangent line will be worked until you get the derivative and then it will be plugged in and solved to have an equation
A regular derivative will be worked to get only the derivative.
Give an example of finding an equation of a tangent line
EXAMPLE:
Find the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6)
Sqrt x-3
Simplifies to (x-3)^1/2
Take the derivative
d/dx= 1/2( x-3)^-1/2 (1)
convert back to a fraction 1/ 2(x-3)^1/2
plug in for x 1/ 2 (39-3)^1/2
Solve = 1/ 2 (36)^1/2
Simplify= 1/ 2 sqrt 36
Solve= 1/12
Therefore the derivative of f(x)= sqrt x-3 at the point (39,6) is 1/12
Because this is the equation of a tangent line you must plug into the equation of a line formula to answer the question of the problem.
Therefore you will plug into y-y1 = m(x-x1)
plug in= y-6 = 1/12 (x-39)
solve: y-6 = 1/12(x-39)
+6 +6
Y= 1/12x + 11/4
Therefore the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6) is y=1/12x + 11/4
and an example of finding a regular derivative
EXAMPLE:
Find the derivative of the function f(x)= sqrt x-3
Sqrt x-3
Simplifies to (x-3)^1/2
Take the derivative
d/dx= 1/2( x-3)^-1/2 (1)
convert back to a fraction 1/ 2(x-3)^1/2
Therefore the derivative of f(x)= sqrt x-3 is 1/ 2(x-3)^1/2
And because this problem calls only for the derivative of sqrt x-3 you will stop working here and your answer is
D/dx= 1/ 2(x-3)^1/2
To find the equation of the tangent line, you would do the same thing you do when you are asked to take the derivative, EXCEPT for one thing. When you are asked to find the SLOPE of the tangent line, all that means is to take the derivative and then you're done. However, when they ask you for an EQUATION, you still have to take the derivative first, but you don't stop there. They usually give you a point for those problems, and if not you would have to find the x and y coordinates yourself. In order to get the equation, you must first plug the x-coordinate into the derivative to get your slope (m). Then you plug your point and slope into the point-slope formula y-y1=m(x-x1). And that equation would be your answer. But sometimes you may have to simplify that equation if they want it in y=mx+b form or standard form which is Ax+By=C. *The difference between an equation of a tangent line and a regular derivative is that for one you have to have an equation like y= "whatever"..and for a regular derivative your answer could simply be a number or some type of polynomial (and there would not be an equal sign anywhere).
ReplyDeleteSo here are some examples:
1.) Find the equation of the tangent line (in slope-intercept form) when
f(x) = (x+5)/(x-5) at the point (4,5).
*So the first thing you have to do with this is take the derivative of f(x). And to do that you have to use the quotient rule
*So for the first step of the quotient rule you get: (x-5)(1)-[(x+5)(1)]/(x-5)^2
*Simplifying that^ you get:
x-5-x-5/(x-5)^2
*You can cancel out the x's which leaves you with
-10/(x-5)^2 and that's your derivative function
*Since they want an equation, the next thing you have to do is find the slope. To do that you have to plug the x-coordinate into the derivative you just got
*So you get: -10/(4-5)^2 = -10/(-1)^2
So m = -10
*Now that you have your slope and you were given a point, you can plug those into the point-slope formula.
*So you get: y-5=-10(x-4)
*Buttttttt, you're not done yet! That equation isn't in slope intercept form so you have to simplify it. *slope intercept form is y=mx+b
*So first you can distribute the -10 through and you get y-5=-10x+40. Then you add the 5 over to 40 on the right side of the equation
*So your final equation is y=-10x+45
2.) y = (x^3-2)^2
^^Find the SLOPE of the tangent line.
*Keeping in mind that the "slope of the tangent line" means ONLY to take the derivative, that's all you have to do here. And in order to take the derivative here, you have to use the chain rule
*So starting with the outermost thing, the exponent, you bring the 2 out in front, recopy the middle, and subtract one from the exponent so that you have:
y'=2(x^3-2)^1
*Now you multiply that^ by the derivative of the inside which would be 3x^2
*So you have y'=2(x^3-2)(3x^2)
*Simplifying that you get 6x^2(x^3-2) and that's your final answer.
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ReplyDeleteWhen the directions ask to find the equation of a tangent line, it is different then just finding a regular derivative. Don’t forget that the derivative is the slope of a tangent line; therefore, in order to find the equation of a tangent line, you need to take it one step further. You first must take the derivative of the given equation. Once you’ve done that, plug in the x-value of the given point into the derivative to get a slope. Then, plug the slope and point into the point-slope formula:
ReplyDeletey – y1 = m(x – x1)
Now you have your equation!
Examples of regular derivatives:
1) Find the slope of the tangent line f(x) = x^2 + 5.
f ’(x) = 2x
2) Find the derivative of the following equation: 2x^3 – 6x + 9
d/dx = 6x^2 – 6
Examples of finding equations of tangent lines:
3) Find the equation of the tangent line f(x) = x^3 at (2,1).
f ’(x) = 3x^2 = 3(2)^2 = 3(4) = 12
m = 12
y – 1 = 12(x – 2)
*Some problems may want you to simplify the answer further by solving for y and putting the equation in slope-intercept form.
4) Find an equation to the tangent line for the graph of f at the given point:
f(x) = 5x^2 + 2x , (4,2)
f '(x) = 10x + 2 = 10(4) + 2 = 42
m = 42
y – 4 = 42(x – 2)
this is my second time doin this stupid thing because of my stupid computer... so the derivative is just the slope of a tangent line and in order to find the equation for a tangent line you have to plug the derivative of the equation into the equation of the line: y - y1 = m(x - x1)
ReplyDeletefor example: find the equation of a tangent line with f (x) = 2x^4 at (2,3).
f '(x) = 2(8)
m = 64
y - 3 = 64 (x - 2)
How do you find the equation of a tangent line? To find the equation of any line there are two things you need: a point and a slope. To find a slope you just find the derivative of the equation given. To find your point, if it isn't given, you plug in your derivative back into the original equation to get a y value. Then find your x afterward.
ReplyDeleteEx. Find the derivative of x^2+2
Using the power rule and constant rule you can simply find the answer is 2x.
Find the equation of the tangent line to x^2+2 through the point (4,4)
You start off the same way by finding your derivative, which is 2x. Then to get a slope we plug in the given x value to the derivative, which gives us 2(4)=8. The slope is 8. Now we have a point and a slope, which we can plug into point slope form.
y-y1=m(x-x1) where m is our slope
Final answer: y-4=8(x-4)
This week’s prompt is on finding an equation of a tangent line, and how it is differs from finding a regular derivative. **Now the problem can ask to find the equation of a tangent line, or the slope of a tangent line. They are both worked the same way, but with slope you go further into working out the problem. They might give you an equation in standard form and ask you to find the equation of a tangent line, or they might give you a point, or sometimes both. *If you are just looking for the equation of a tangent line. To find the equation of a line you would have to find the derivative, once that’s done you would plug into point slope form which is y-y1=m(x-x1) <-M being slope. Use the point it gives you to plug in as well. *Now if you are finding the slope of a line it is a little more complicating. You would take the derivative, and then take the slope from the standard form it gives you. Set the slope and the derivative equal, plug back into one of the equations it gives you then put it in point slope form.
ReplyDeleteExample:
y = 3x + 4, (2,10)
-Starting off you can automatically put it in point slope form
y - 10 = 3(x - 2)
**Notice if you solve the equation
y - 10 = 3(x - 2)
y - 10 = 3x - 6
+10 +10
y = 3x + 4
Now for a regular derivative you would simply solve it and once your done you would leave it as it is. Do not plug anything in, or put it in any kind of form, or equation.
Example:
X^2+2x^3
2x+6x^2 <-Final Answer
How do you find the equation of a tangent line?
ReplyDeleteFirst you need a point, and then a slope. The slope of a tangent line is the derivative of the function, then plug in your point. If it does not give you a point, plug the derivative into the equation.
Dy/Dx x^3 - 4
by using the power rule this comes out as 3x^2
Find the equation of the line tangent through the point (1,2)
take the derivate (3x^2), plug in your x value which gives you 3(1)^2=3
the equation is y-1=3(x-1)
how do you find the equation of a tangent line? to start, two things are needed; a slope and a point. the slope is equal to the derivative of the original equation, and to find the point, all you have to do is plug it in. if theres no point, just plug the derivative straight into the original equation:
ReplyDelete3x^2-5 at (1,3)
take the derivative: 6x
plug in to get slope: 6(1) = 6
put in point slope form: y-y1 = m(x-x1) which equals: y-3 = 6(x-1)
to find the equation of a tangent line, you have to use the derivative formula, which is the limit as &x approaches 0 of the function (f(x+&x) - f(x))/(&x). This is to find the slope of a tangent line. Now to find the equation, you first have to find the slope and have a point, then plug in the slope to the point-slope formula, y-y1=m(x-x1). The regular derivative is just taking the regular derivative, as it says, just an equation.
ReplyDeleteFinding a tangent line:
Find an equation of the tangent line to the graph of f(x) = x^2 when x=-2.
dy/dx = 2x
=2(-2) = -4
y: (-2)^2 = 4
y-4 = -4(x+2)
Find the [regular] derivative of 2x^2.
4x.
Uhh........................I'm just gonna say it straight up...........I have NO clue how to do that..........
ReplyDelete