Question

Anyone good with calc graphing? Review problem i need some help with: Find the equation of the tangent line to the curve y=1⁄x at the point B(2,1⁄2). Can someone walk me through the steps for this?

Answer 1

can you find the 1st derivative..?

Answer 2

y-y0=m(x-x0). y0=1/2, x0=2, m=y'(2)

Answer 3

what i had was y= 1/2x- 1/2 but thats wrong based off that i guess...

Answer 4

ok... thats not quite correct... if you write the problem in index form its

\[y = x^{-1}\]

this is what you need to find the derivative of...

Answer 5

so apply the differentiation rule

\[y = x^n...then...y' = n \times x^{n -1}\]

Answer 6

I guess I'm having issues with my labeling in these words problems then, can we change the integers and do an example walk through so I can see ?

Answer 7

Because it makes no sense what I had down as the n value apparently

Answer 8

ok... so n = -1

\[y' = -1 \times x^{-1 -1} = -x^{-2}\]

does that make sense..?

Answer 9

which can be written as

\[y' = \frac{-1}{x^2}\]

Answer 10

that makes sense to me from prior problem right, for the variable labeling how do you know what exactly n is when given the coordinates and y?

Answer 11

ok... so n is just the power of x.

So what you have from the 1st derivative is the equation for the slope of the tangent.

to get the specific slope at the point (2, 1/2) just substitute x = 2 into the 1st derivative.

what do you get...?

Answer 12

@campbell_st You've always been a smart person! I thank you in a large amount for all the help you do on OpenStudy! I just had to say it.

Answer 13

" for the variable labeling how do you know what exactly n is when given the coordinates and y?"

You're overthinking it. n is the power on x. That's it. Has nothing to do with the points given.

When finding the derivative of y=1⁄x ignore EVERYTHING ELSE, all you need is y=1⁄x. No points or anything.

The use of the point B(2,1⁄2) comes later. You're mixing too many things together, do one step at a time.

Answer 14

so I need to solve first what campbell st initially posted?

Answer 15

y=x^-1 so thats just y= 2^-1 so 1/2 is the 1st?

Answer 16

Yep, when finding the derivative, the ONLY thing you should be looking at is \[\Large y = x^{-1}\] then use the power rule

Answer 17

"y=x^-1 so thats just y= 2^-1 so 1/2 is the 1st?" where did x go? \[\Large y = x^{-1}\] \[\Large y' = -1 x^{(-1 -1)}\]

Answer 18

well not quite

\[y' = \frac{-1}{x^2}...x = 2...gives.... y' = \frac{-1}{2^2}\]

so the slope of the tangent is m = -1/4

does that make sense?

Answer 19

thanks @Mikeyy1992 lots of bright people here who help with maths problems...

Answer 20

I plugged the 2 into for X which was apparently wrong, thats how i got the 1/2. but i understand now the -1/4 what i did wrong yeah

Answer 21

Welcome. @campbell_st Yes, I know that very well. It's just amazing to see people put time in helping others.

Answer 22

i plugged the right value in just wrong format

Answer 23

well the slope at the point x = 2 is m = -1/4

so here is a short cut

the slope intercept form of a straight line is

y = mx + b

in your problem you have the tangent equation as

y = -1/4 x + b

use the point (2, 1/2) and substitute x = 2 and y =1/2 into the tangent equation to find the value of b, and thus the equation of the tangent.

hope this helps

Answer 24

so i had y=-1/4x+ 1 so based off that shortcut im right? since -.5 on the right has to be added to .5 on the left...right?

Answer 25

thats the equation I came up with...

Answer 26

much appreciated

Answer 27

glad to help

just practice derivatives... and good luck