Question

Finding the equation of the line that is tangent to the following curve at (1, pi/2):
it then gives the equation: 2xy+pi sin y=2pi
What rules do i use to find the derivative?

Answer 1

Use the chain rule.

Answer 2

\[
2xy+\pi \sin y=2\pi
\]First do \[
(2xy)'
\]This requires product rule.

Answer 3

Next do \(\pi\sin y\), this just requires chain rule.

Answer 4

\[
2xy+\pi \sin y=2\pi \\
(2xy)'+(\pi \sin y)' = 0
\]

Answer 5

Need help?

Answer 6

sorry! was working on helping someone. one second :D

Answer 7

the deriative of 2xy is 2. but what do i do with 2pi?

Answer 8

Nope \[
(2xy)' = (2x)' \cdot y + 2x\cdot y' = 2y+2xy'
\]

Answer 9

whaaat. ok so, the deriative of (2x)'=2 why does y change things?

Answer 10

product rule!

Answer 11

Okay suppose \(f(x) = 2x\) and \(g(x) = y\)

Answer 12

I'm just saying \[
(fg)' = f'g+fg'
\]

Answer 13

duh. im stupid.

Answer 14

Now can you do \[
(\pi \sin y)'
\]

Answer 15

yay thank you. It might take me a few minutes. Id be grateful if you checked back :D

Answer 16

I'm not going to check your work unless you show me the steps though.

Answer 17

Of course.
\[\frac{ d }{ dy }(\pi \sin (y))= \pi(\frac{ d }{dy })\sin(y)= \pi(\frac{ d }{ dy })\cos(y)= \pi \cos(y)\]

Answer 18

But you want to differentiate with respect to \(x\)

Answer 19

I have no idea how to do that.

Answer 20

use chain rule: \[
\frac{d}{dx}f(y) = f'(y)y'(x)
\]

Answer 21

Basically\[
\frac{d}{dx}f(y) = \frac{df}{dy}\frac{dy}{dx} = f'(y)y'
\]

Answer 22

You found \(f'(y)\) already.

Answer 23

\[
f'(y) = \pi \cos y
\]

Answer 24

oh ok i thought you were asking me to use the chain rule to find the derivative of pi sin y lol

Answer 25

was trying to wrap my head around that

Answer 26

\[
\frac{d}{dx}\pi \sin y = (\pi \cos y)y'
\]

Answer 27

so \[\pi \cos(y)\] is correct then?

Answer 28

Notice the \(y'\)

Answer 29

ok. i see it. why is it there? :(

Answer 30

Okay.... Look chain rule says: \[
[f(g(x)) ]' = f'(g(x)) g'(x)
\]

Answer 31

Let \(y = g(x)\) and you get: \[
\frac{d}{dx}f(y) = f'(y) y'
\]
In this case \( f(y) = \pi \sin y \)

Answer 32

is the \[f'= 2y+2xy'?\]

Answer 33

... I'm tired dude...

Answer 34

lol. sorry. its fine. thank you for your help.

Answer 35

I can show you my answer...

Answer 36

no thats okay. I'd rather figure it out first.

Answer 37

I dont understand how to apply the chain rule to different equations, im doing practice problems to attempt to understand it better, But i appreciate the effort!

Answer 38

\[
2xy+\pi \sin y=2\pi \\
(2xy)'+(\pi \sin y)' = 0 \\
2y+2xy'+(\pi \cos y)y' = 0 \\
(2x+\pi \cos y)y' = -2y \\
y' = -\frac{2y}{2x+\pi \cos y}
\]