Find the derivative of the function y=2 x^2-13 x+5 and use it to find an equation of the line ta

Find the derivative of the function y=2 x^2-13 x+5 and use...

Key Concepts

Derivative

In calculus, the derivative of a function represents the rate at which the function's value changes with respect to changes in its input. It provides the slope of the curve at any given point, making it a fundamental tool in understanding the behavior of functions.

Power Rule

The power rule is a basic differentiation technique used for finding the derivative of a term in the form of x raised to a constant power. It states that for any function f(x) = x^n, the derivative is f?'(x) = n*x^(n-1), which enables the straightforward differentiation of polynomials.

Tangent Line

The tangent line to a curve at a specific point is a straight line that just touches the curve at that point and has the same slope as the curve there. Determining the tangent line involves computing the derivative at the point of interest to obtain the slope and then using the point-slope form to write the equation of the line.

Point-Slope Form

The point-slope form is a method for expressing the equation of a line when you know a point on the line and its slope. It is written as y - y1 = m(x - x1), where m is the slope and (x1, y1) is a known point on the line. This form is particularly useful for finding the tangent line once the slope and point of contact are known.

Transcript

00:03 Hello everybody, so today our question is find the derivative of the function y equals 2x squared minus 13 x plus 5 and use it to find an equation of the line tangent to the curve at x equals 3.

00:17 Okay, so we'll be solving this question two parts, finding derivative and then finding the tangent line.

00:23 We'll find the derivative by using our limit definition of the derivative and before we start it's much easier if we reframe this equation in terms of the function f of x be helpful for our limit definition which by the way our limit definition of the derivative would be f of x or i'm sorry f prime of x which is a derivative of f is going to be equal to the limit as h approaches zero of f of x plus h minus f of x all over h okay so this uh little definition is what we're using to find the derivative however since our function was given in terms of y um let's also rewrite this as terms of d y over d x or more uh notations that match uh what the problem was given to us okay so this is we're going to be using to solve the problem so let's actually apply it so for this specific case we're going to be substituting f of x the equation right here for f x to our limit definition so following that it's going to be equal to the limit as h approaches zero and actually to break this up to make it easier let's do the f of x plus h in red and the f of x plus h in red and the f of x in blue just make it easier to look on the eyes because these are going to be pretty long terms when we're writing out first so again this part right here it's going to be in red and the other little bit blue so we actually have two um x plus h squared so we have the f plus h and the subsection the x right here minus 13 of x plus h then plus five okay and then the minus is black that's neutral minus f of x which we'll be doing in blue so this part will be done in blue we'll actually bracket it to show that it's literally just f x so it's just 2x squared minus 13 x plus 5 okay then we're going to have all of this divided by, oh, white, no, black again.

03:04 All this, divided by h.

03:07 Okay, so pretty long.

03:10 Let's just expand this out, so it would be easier for canceling out terms.

03:14 So it's going to be the limit as h pros to zero of 2x squared, plus 4xh plus 2h squared.

03:27 And expanding this out we have minus 13x minus 13 h plus 5 and then we're going to distribute the the negative to this term right here these terms so it will be actually not oh wait plus 5 i forgot that so it'll be minus 2x squared and the negative negative turns to a plus 13 x minus a 5.

04:00 Okay, then all this is over h again.

04:05 Okay, so this is pretty, pretty long limit.

04:09 But luckily, we can cancel things out.

04:12 It's fun.

04:12 We can make things shorter.

04:13 We like canceling things out.

04:15 So let's do this in green.

04:19 Okay, so doing this in green, let's see what terms cancel out.

04:22 I see a 2x squared here, and i see a negative 2x squared here.

04:27 Those cancel out, which is good.

04:30 Then 4xh, no, that's the only thing here.

04:34 2h squared, that is also very unique.

04:37 Negative 13x.

04:39 I see a positive 13x.

04:41 Great.

04:42 Cancel that out.

04:43 Plus 5, negative 5.

04:45 Good.

04:46 Okay, so this very large limit is actually going to shrink very much to our liking.

04:54 So we just have the limit as h to put to zero of, of 4xh plus 2h squared minus 13h.

05:05 All that's still over h.

05:08 And what do you know? there's h in every turn on top.

05:10 We have h on bottom, which will cancel out.

05:13 So then this will leave the limit as h approaches 0 of 4x plus 2h minus 13.

05:24 And actually if you think about it, 17 to think h into this limit won't really cause any, won't break any rules of math because there's subducing zero there, it does not become indefinite.

05:36 So that's what we're going to do.

05:38 So then this whole limit itself, we can drop the limit sign and it becomes 4x plus 2 .0, i mean, 10, minus 13.

05:48 Then we just have 4x minus 13, which is what we wanted.

05:54 So, are d y over d x is actually just equal to 4x minus 13 which is great okay that's the first part of the problem second part of the problem is we're going to be finding the equation to the tangent line of the curve at x equals 3 let's do that on another page okay so in order to find this tangent line equation um let's use a point slope form so we know it's going to be a linear line because tangent lines are straight lines...

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