Question

Find dy/dx by implicit differentiation.
(sin πx + cos πy)^4 = 67
dy/dx=?

Answer 1

(πcos(x)-πy'sin(πy))(4(sinπx + cosπy)^3)=0

Can you take it from there?

Answer 2

I first did the power rule to get 4(sin πx + cos πx)^3
then I multiplied by the derivative of the inside, π(cos πx - sin πx(dy/dx))
All that equals 0 and I'm not sure what to do now.

Answer 3

I made two typos.. its supposed to be 4(sin πx + cos πy)^3 and π(cos πx - sin πy(dy/dx))

Answer 4

What did you do here: (πcos(x)-πy'sin(πy))?
I follow the π's but why was the π removed from cos(πx)?

Answer 5

What pi outside the sin?

Answer 6

Sorry about that, I seem to be thinking of a typical derivative. Your answer is correct, just make sure you factor the variables out to find y'

Answer 7

How can I get the y' by itself thought? theres nothing on the ride side to divide into, 0/anything = 0

Answer 8

Alright, the 4(sin πx + cos πx)^3 would be taken out because 0/x=0

Next, I would distribute the pi and add the part with y' to the other side

Now you just need to separate the y' from the equation, so divide both sides by πsin(πy)

Answer 9

So something along the lines of -cos(πx)/sin(πy)=dy/dx would be my answer? Not to be rube but I would think canceling out the 4(sin πx + cos πy)^3 would screw something up?

Answer 10

Yes, that would be correct. You can treat 4(sin πx + cos πy)^3 as a variable, say k, so
k=4(sin πx + cos πy)^3

We know that when we divide zero by any number we end up with 0, so 0/k=0 as well

Answer 11

Hmm I see, This is for an online portion of calculus so I can check my answers but it still says its wrong lol thanks for the help though.

Answer 12

Try y'=cos(πx)csc(πy)

Answer 13

Okay, well I found another way to do it without diving by zero but the answers came out to the same and it did work, not sure why it didn't accept it the first time. Maybe I typed it wrong but cos(πx)/sin(πy) is correct. Thanks.

Answer 14

yw

Answer 15

\[
4 (\cos (\pi y(x))+\sin (\pi x))^3
\left(\pi \cos (\pi x)-\pi y'(x)
\sin (\pi y(x))\right)=0\\

\pi \cos (\pi x)-\pi y'(x) \sin (\pi
y(x))=0\\
y'(x)=\cos (\pi x) \csc (\pi y(x))
\]