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Finding The Equation Of A Tangent Line & Calculating A Derivative using the defenition of A Derivative by: Lazaro Reyes.

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Presentation on theme: "Finding The Equation Of A Tangent Line & Calculating A Derivative using the defenition of A Derivative by: Lazaro Reyes."— Presentation transcript:

1 Finding The Equation Of A Tangent Line & Calculating A Derivative using the defenition of A Derivative by: Lazaro Reyes

2 Calculating a derivative using the defenition of a derivitive Before finding the equation of a tangent line we must first understand Derivative. I will show and explain problems for better understanding. The definition that i find better to work with is the following. It basically says that f prime of x = the limit as h goes to 0, f of x+h- f of x over h.

3 Let us find the derivative of the quantity.........f(x)=12+7x Every where theres x replace it with x+h. so the first part looks like this 12+7(x+h) Then you subtract it by the original quantity (12+7x) In this case h will equal 0 Then you will multiply it out. so it becomes 12+7x+7h-7x over h. Then you cancel all the like terms and you are left with 7h over h Cancel h to leave 7 alone and 7 becomes your answer.

4 Lets try some more problems........G(t)=4t over t+1 Using the definition of derivative says that we must replace t for (t+h). After you have done that your equation will look like this 4(t+h) over t+h+1. Then you subtracted by its original equation and divide by h. Then you look for the common denomanator to simplify.so the equation will now look like this... After doing that you will cancel all the like terms. So you will end up with 4h over (t+h+1) (t+1). Get rid of h you will have to change h to h over 1 and then put one on top of the h so it can allow it to cacel it out. After doing this you will end up with G prime of (t) = 4 over (t+1) (t+1) which can be equal to 4 over (t+1)squared.

5 Finding the equations of Tangent Lines One thing that is really necessary is to know the point-slope formula which is y- y1=m(x-x1). The other thing is derivative. In the equations i will demonstrate, you will see the steps required to obtain the equation of the Tangent Lines.

6 Finding the Equation of a Tangent Line at points (2,4) on y=x squared The point-slope formula is needed for these type of problems............ y - y1= m (x-x1) You have your two points (2,4) so keep that in mind. Now, the derivative of x squared is 2x and we are getting this from y= x squared. Since we already know our x point or coordinate.we plug it in for every x. After doing this you will get 4 and that is your slope. Now we take our point slope formula and we plug in what we already know. Soon your equation will look like this : y - 4=4 (x-2) Then you solve it first on the right side. you will get y-4=4x-8 so you add the for from the left to the right leaving y alone when you solve it piece by piece you will soon get your answer y=4x-4

7 The problem continue... Again, your points are (1,9) and this will be on f (x)=(1+2x)squared. something to keep in mind is that the derivative of 1 is 0 and de derivative of positive 2x is positive 2. To start of we will set the equation to f prime of x =2(1+2x)to the power of 1 (2). because you take the power and put it in front and the inside you leave it all to the power of 1 and we saide the derivative of 2x is 2 so we put that after. Then you multiply two by the outside 2 which gives you 4 (1+2x) Then we plug in what we know x = 1 Then you solve thee inside first which will give us 3 Then multiply and we get 12 so y - 9=12 (x-1) and we could put it in standard form if asked...

8 Thank You for listening...


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