Question

Find the tangent line to the curve \[y=\cos(x+y)\] at the point \[(\Pi/2,0)\]

Answer 1

You need to find dy/dx of the curve

Answer 2

Do you have any trouble differentiating on both sides?
(chain rule twice on the right side - once for the part inside the cosine - second chain for the y (the y') )

Answer 3

Ok, what is the derivative of y?

Answer 4

Would it be y'?

Answer 5

yes exactly

Answer 6

So it's -sin(x+y) (y')

Answer 7

y is a function of x, (just as f(x) is)

The derivative of f(x) is f'(x), AND
the derivative of y is y'.

Answer 8

the derivative of x is 1, not 0

Answer 9

(Nice username btw, I like that)

Answer 10

Oh okay, I forgot the 1.
\[-\sin(x+y)\times(1+y')=f'(x)\]

Answer 11

yes,

y' = -sin(x+y) × (1+y')

Answer 12

Thanks!

Answer 13

Ok, now you need to isolate the y'

Answer 14

(y' is same as dy/dx)

Answer 15

\(\large\color{#000000 }{ \displaystyle y' = -\sin(x+y) \times (1+y') }\)

\(\large\color{#000000 }{ \displaystyle y' = -\sin(x+y) -y'\sin(x+y) }\)

Answer 16

\(\large\color{#000000 }{ \displaystyle y' +y'\sin(x+y)= -\sin(x+y) }\)

Answer 17

\(\large\color{#000000 }{ \displaystyle y' (1+\sin(x+y))= -\sin(x+y) }\)

\(\large\color{#000000 }{ \displaystyle y' =\frac{ -\sin(x+y) }{1+\sin(x+y)} }\)

Answer 18

I can't read that code clearly, what is the point of tangency? )

Answer 19

\((\pi/2~,~0)\) that?

Answer 20

yeah

Answer 21

So, plug that point \(x=\pi/2\) and \(y=0\) into the derivative, and see what you get.

Answer 22

x=-1/2

Answer 23

x ?

Answer 24

you mean the result is -1/2?

Answer 25

Oh I meant y'

Answer 26

yeah :)

Answer 27

-1/(1+1)=-1/2
(because sin(π/2)=1)

Answer 28

So your slope is -1/2.
And the point is \((\pi/2,0)\)

you got it from here

Answer 29

Ok, thank you very much!!

Answer 30

If you would like to later confirm the answer, then,
here is the graph of the curve, the point, and the tangent line.
https://www.desmos.com/calculator/hdtir0corc

Answer 31

Alright, thanks! :)

Answer 32

Good Luck!