Question

Find the equation for the tangent line of the graph y = sinh(1-x^2) for points (1,0)

Answer 1

ewww, sinh

Answer 2

we got a formula for that one?

Answer 3

sinh(x) = (e^-x - e^ix)/2 or something like thjat

Answer 4

lol, that was simple enough; D[sinh] = cosh

Answer 5

haha yeah sinh = cosh sinh = e^x - e^-x/2

Answer 6

the eq of the tan line is just:
\[tan_y=f'(1)x-f'(1)1+0\]

Answer 7

\[f' = -2x\ cosh(1-x^2)\]

Answer 8

f'(1) = -2 cosh(0) whatever that maybe

Answer 9

so all we do is plug in the points 1 and 0... after taking the derivative?

Answer 10

well, the point is what we use to calibrate the line itself; and since x=1 is our value for x, we plug that into the f'(x) to get the value of the slope at (1,0)

Answer 11

so yeah, slope = f'

slope at x=1 is f'(1)

Answer 12

okay... So it could be any number, it just needs to create the line it is asked?

Answer 13

correct

Answer 14

If they had given a different point, say (3,-7) then slope = f'(3) and the rest of the line equation from algrebra forms as it forms

Answer 15

okay, so we only use the x value then, ic. So why is the formula f(1)x - f(1)1 - 0? Where do we use that?

Answer 16

recall the point slope form of a line?

y -Py = m(x-Px)

y = mx -mPx + Py ; (Px,Py)

Answer 17

m=f', (Px, Py) is given as the point

Answer 18

ah okay so we need to find m, and then have our equation. so it's y - 0= m (x-1)?

Answer 19

correct; and m = f'(1) in this case

Answer 20

derivative is equation for slope at any given point

Answer 21

okay so for the 3,7 point f(3) = m?

Answer 22

f'(3)

Answer 23

correct

Answer 24

okay so in this case it would be y-0 = cosh(0) [1] (x-1) y = 1(x-1) = 1x-1

Answer 25

f' = -2x cosh(1-x^2) ; when x=1; f'(1) = -2cosh(0)

Answer 26

cosh(0) = 1 so f'(1) = -2

Answer 27

oh boo I thought I was cool and didn't check the answer and guessed haha... :(

Answer 28

okay so y = -2x +2

Answer 29

tany = -2x+2(1)+0

tany = -2x+2 ; correct

Answer 30

ah so that's where that formula comes in :)

Answer 31

yeah, i find it simper to work than the usual point slope form that you have to then work around to get it into this form anyways

Answer 32

as long as you can test it :)

Answer 33

tany=f′(1)x−f′(1)1+0 One o more thing on this, if it was the point (3,7) would it be tany=f′(3)x−f′(3)1+0

Answer 34

no; IF the point had been (3,7) we would have used this point instead.

tany = f'(3)x -f'(3)3 + 7

Answer 35

so f'(x)x - f'(x)x + 7?

Answer 36

except the 2nd x is a constant.

Answer 37

+ y

Answer 38

GRR this test isn't going to go well :'(

Answer 39

close, except for the equation of the tangent line use a specific value associated with the derivative; so instead of confusing xs, I use Px and Py for the pointX component and pointY component

tany = f'(Px)x - f'(Px)Px + Py

Answer 40

you might be used to subscripts for your given point and generic for the unknowns

\[y_{tan}=f'(x_o)x-f'(x_o)+y_0\]

Answer 41

or even:
\[y_{tan}=f'(c)x-f'(c)c+f(c)\]

Answer 42

such that the point given is: (c,f(c))

Answer 43

tany = f'(Px)x - f'(Px)Px + Py works :)