Question

Chain Rule •
"Find an equation of the tangent line at the given point."

LaTeX below (2 questions).

Answer 1

\(\Large{\color{seagreen}{\text{A.}}}\) \(\large{y=3^{x^{2}}}\) at \((2,81)\)
What I have so far:\[y'=(3^{x^{2}})'=(3^{u})'\times(x^{2})'=(3^{u}\ln{u})\times(2x)=(3^{x^{2}}\ln{x^{2}})\times(2x)\]

\(\Large{\color{royalblue}{\text{B.}}}\) \(\large{y=\cos{\sin{x}}}\) at \((0,1)\)
What I have so far:\[y'=(\cos{\sin{x}})'=(\cos{u})'\times(\sin{x})'=-\sin{(\sin{x})}\times\cos{x}\]

Answer 2

Woops, \(\large\rm \left(3^x\right)'=3^x(ln3)\)

Log of the base.

Answer 3

URGH! I made a typo.
Sorry @zepdrix lol

Answer 4

Is it good otherwise, and is it necessary for me to proceed further?

Answer 5

Well, yes, you've only finished about half of the problem so far :)
Remember that the derivative function gives us `slope of our tangent line`.
It doesn't give us the entire tangent line, only the slope.

Answer 6

Oh, I meant differential-wise haha. I want to make sure I'm solid enough to calculate the slope :-)

Answer 7

Mmmm yes, looks good besides that!
The way you wrote the function in part B is rather confusing.
Use some brackets. cos(sin x).
Derivative turned out well though.

Answer 8

Er, yes. I keep forgetting that in LaTeX lol

Answer 9

Alright, so I can start calculating the tangent slope in both?

Answer 10

Sure.

Answer 11

Okay, well...
For (B) I got this... when x=0,\[-\sin{(\sin{0})}\times\cos{0}=-\sin{0}\times1=0\]

Answer 12

need help

Answer 13

what do you need help on

Answer 14

ok

Answer 15

you can wait

Answer 16

First solve for y as a function of x) and implicitly. 8. y2+2xy+4 = 0 at (-2, 2). In exercises 9-20, find the derivative y' (x) implicitly. In exercises 21-28, find an equation of the tangent line at the given point.

Answer 17

For part B you got m=0?
Ok that sounds right!

Answer 18

I have no idea what you are doing. @bonnieisflash1.0

Okay, thank you @zepdrix
Except I am not sure how to find the equation of the tangent now. Would it be a horizontal line?

Answer 19

If the slope is zero, then yes a horizontal line :)\[\large\rm y-y_o=m(x-x_o)\]Plugging in the point (0,1) and m=0,\[\large\rm y-1=0(x-0)\]

Answer 20

Ah, I see. Thank you for clarifying @zepdrix :-)

Answer 21

I got a very weird answer when putting the value \(x=2\) into the derived equation \((3^{x^{2}}\ln{3})\times2x\). It seems to be an overly large number... @zepdrix http://www.wolframalpha.com/input/?i=3%5E(4)ln(3)*4

Answer 22

\[\large{(3^{x^{2}}\ln{3})\times2x\rightarrow(3^{(2)^{2}}\ln{3})\times2(2)}\approx355.95...?\]

Answer 23

It's difficult to see because the line is very very steep,
https://www.desmos.com/calculator/zy17hydp3t
But this slope value appears to be accurate, yes?

I scaled the axes a bit so it's easier to see.
3^(x^2) is extreeeeeemely steep around x=2.

Answer 24

Hmm, yes, it does seem very steep...

WAIT, how did you do that table on Desmos?! D:

Answer 25

umm :d

Answer 26

I dunno :d one of the buttons .. somewhere .. on there

Answer 27

D:
...so the slope is basically \(\approx356\)?

Answer 28

Ya something like that, 4(ln3)3^4

Answer 29

Alright thank you :-) @zepdrix