Question

find y'
X^4sec(y^3)-2x=y^2cos3x+11

Answer 1

Differential Equation?

Answer 2

yea

Answer 3

i know to use product rule and implicit, but i jus need help solving

Answer 4

Lets just verify your equation
\[x^4Sec(y^3)-2x=y^2\cos(3x)+11\]
?

Answer 5

yea thats the equations

Answer 6

I think it's a good idea to do this implicitly instead of trying to solve for y. If you can solve for y, then you're some kind of magician.

@anumrox try to use the product rule and do it implicitly like you say. If you mess up, it'll be perfect because we'll just see exactly what we need to see to help correct you. So give it your best shot and we'll help. =)

Answer 7

\[y'=x ^{4}(secy^3tany^3)\times3yy'-\cos3x \times2yy'=3y^2\sin3x-4x^3secy^3\]

Answer 8

my final answer is
\[y'=\frac{ -3y^2\sin3x-4x^3secy^3 }{ 3yx^4(secy^3tany^3)-2ycos3x }\]

Answer 9

what do you guys think?

Answer 10

I think you are on the right track, but your first step is not right. This is sort of a habit everyone falls into when they start doing implicit derivatives, and that is that when you take the derivative, the left side is not equal to y'.

I'll show you an example, \[\large y=x^2\] We know the answer, but let's just do some algebra to rearrange it without changing the relationship between x and y. This will keep the derivative the same.
\[y^{1/2}=x\]Take the implicit derivative of this and you will get: \[\large \dfrac{1}{2}y^{-1/2}y' = 1\] Now all we did was the chain rule, and so we can solve for y' now. Notice that the whole thing isn't equal to y'. In fact, if we use the other notation it might become clearer that we're doing the same thing to both sides. \[\large \frac{d}{dx}(y^{1/2})=\frac{d}{dx}(x^1)\]\[\large \frac{1}{2}y^{-1/2} \frac{dy}{dx}=1x^0\frac{dx}{dx}\] See how we did the chain rule on both sides? We multiply by the exponent, subtract 1, and then multiply by the derivative on the "inside". But notice in the case of x, we have x^0 which anything to the zero power is 1. And what's dx/dx? It's the change in x with respect to x. Well everything we move x we move x by the same amount, there's a 1 to 1 relationship there because they're the same thing! So dx/dx=1. Fairly straight forward?

Now if you solve for y' from this, \[\large \dfrac{1}{2}y^{-1/2}y' = 1\] we get \[\large y' = 2y^{1/2}\] and if you look above, remember, y^1/2=x so we plug that in \[\large y' = 2y^{1/2}=2x\] Hey! That's the same answer as if we didn't do it implicitly. This is a good sign, since we got here a different way and it's still consistent.

So maybe this helps explain how to do this? I know it's kind of a lot to see at once.

Answer 11

I don't normally like to do this, but I am going to put the answer here because I'm talking with someone about what the answer should be through private message and I think I might as well just put it out here. \[\large \frac{d}{dx}[x^4 \sec (y^3)-2x]=\frac{d}{dx}[ y^2\cos(3x)+11] \\ \large 4x^3 \sec (y^3)+x^4 \sec (y^3) \tan (y^3) (3y^2) y'-2= 2y y' \cos(3x)- y^23\sin(3x)\]
I hope this clears up what should happen. I can answer any questions about any of this if you are confused. Try to follow along as best you can, I used the product and chain rules.