Question

No idea how to do this question. :(

Answer 1

Find dy/dx at the given point.
\[\cos^{2}(9y)=x+y\]

Answer 2

pts: \[(\frac{ 2-\pi }{ 4 }) (\frac{ \pi }{ 4 })\]

Answer 3

do you know how to do implicit differentiation?

Answer 4

yes, that's how i need to start the question.
i had: \[-\sin^{2}(9y)=1+y'\]

Answer 5

opps, there's a 9y' after the sin

Answer 6

\[-\sin^{2}(9y)*9y'=1+y'\]

Answer 7

i dont think the derivative of cos^2 is -sin^2...you have to use the chain rule or the product rule on that

Answer 8

the derivative of cos = -sin

Answer 9

yes. but just because that's true doesnt mean when you square them its true
d/dx (cos x)^2 using the chain rule, take derivative of the external, times derivative of the internal
= 2 (cos x)(-sin x)
=-2sinx cos x
= - sin(2x)

Answer 10

its not cos(x)^2 thought, it's (cosx)^2

Answer 11

if i did it as cos(x^2) it would be -sin(x^2)(2x)
since its (cos(x))^2, use the chain rule
2cosx(-sinx)

Answer 12

what should my y' value be?

Answer 13

\[\cos^{2}(x) = \cos^{2}(x)*-2\sin(x)\]

Answer 14

???

Answer 15

:S

Answer 16

im confused now, you have two different things wrote down for the derivative of cos^{2}(9y)

Answer 17

that some albert einstine questions

Answer 18

it's calculus ....

Answer 19

not 4 me wat are u seanior

Answer 20

theres 2 different things but theyre equivalent

Answer 21

i'm in university ..
and i don't know what one to use now ... im confused.

Answer 22

hold on...im confused too...its been awhile since ive done this

Answer 23

\[\cos^{2}(9y)=x+y\]

Answer 24

i don't think you need to use the chain rule for the cos part.

Answer 25

i know for sure you do...but let me see if i can get this first before i try to explain it to you

Answer 26

okay..

Answer 27

\[\cos ^{2}9y=x+y\]
derivative:
\[2\cos(9y)*-\sin(9y)*9\frac{ dy }{dx } = 1+\frac{ dy }{ dx }\]
you need to use the chain rule to find the derivative of cos^2(9y)
do you see how i did that?
theres 3 functions combined into each other on the left side. in terms of x they would be: x^2, cos(x) and 9x
all of these are combined to get cos^2(9y)

Answer 28

i see, i see.

Answer 29

so now what do you do with the dy/dx

Answer 30

you have to move the one on the right to the left, and then you can factor it out, and then divide by whats left after factoring (the stuff thats not dy/dx)
first, lets clean up the left side a little bit
\[-18\cos(9y)\sin(9y)\frac{ dy }{ dx }=1+\frac{ dy }{ dx }\]
if we subtract dy/dx from both sides, then factor it out of the left side, we get:\[\frac{ dy }{ dx }(-18\cos(9y)\sin(9y)-1)=1\]
then just divide both sides by that big messy term on the left

Answer 31

so that's what y' equals? when i divide both sides? it'll be 1 over the right hand side.

Answer 32

yes

Answer 33

then, you just plug in the point (note that the x coordinate of the point doesnt matter because theres no x in the derivative

Answer 34

\[\frac{ 1 }{ -18\cos(9\pi/4)\sin(9\pi/4)-1 }\]

Answer 35

as long as i did my work right, thats the answer...looks like it can be simplified tho because cos(9pi/4) and sin(9pi/4) are on the unit circle and are easy to find

Answer 36

is cos9pi/4 = root2/2?

Answer 37

same for sin ???

Answer 38

is 2/7 the final answer?

Answer 39

i got -1/10
let me check again tho

Answer 40

ok.

Answer 41

let me know when you re-check.

Answer 42

if sin(9pi/4) = sqrt(2)/ 2 and cos(9pi/4) is the same, -18(sqrt2)(sqrt2)/(2*2)-1 = 10