if y^6=tanh^-1(tan^-4(4x^2+y^2)^2)^9
find dy/dx
show work....
this is just stupid.
y^6 = tanh^-1(u)^9
6y^5 y' = 9 tanh^-1(u)^8 * 1/(1-u^2) du/dx
where u = tan^-4(v^2)
and v = 4x^2+y^2
= cot^4(v^2)
du/dx = 4 cot^3(v^2) (-csc^2(v^2)) (2v dv/dx)
dv/dx = 8x + 2yy'
mash all that together and solve for y'
I see little to be gained from such an artificial jumble.
Here's what wolframalpha.com ends up with:
http://www.wolframalpha.com/input/?i=derivative+y%5E6%3Dtanh%5E-1(tan%5E-4(4x%5E2%2By%5E2)%5E2)%5E9
That's as you typed it. I interpreted it as
http://www.wolframalpha.com/input/?i=derivative+y%5E6%3Dtanh%5E-1(tan%5E-4((4x%5E2%2By%5E2)%5E2))%5E9
which is even messier
To find dy/dx, we need to differentiate both sides of the equation with respect to x. Let's break down the equation step by step:
First, differentiate the left side of the equation using the chain rule:
d/dx (y^6) = 6y^5 * dy/dx
Next, let's differentiate the right side of the equation. We have a composition of multiple functions, so we need to apply the chain rule multiple times. Let's start from the inside and work our way out:
1. Differentiate tanh^-1(tan^-4(u)):
Using the chain rule, we have d/dx (tanh^-1(u)) = 1 / (1 - u^2) * du/dx
Here, u = tan^-4(4x^2 + y^2)^2, so we have:
d/dx (tanh^-1(tan^-4(4x^2 + y^2)^2)) = 1 / (1 - (tan^-4(4x^2 + y^2)^2)^2) * d/dx(tan^-4(4x^2 + y^2)^2)
2. Differentiate tan^-4(v):
Using the chain rule, we have d/dx (tan^-4(v)) = -4 / (1 + v^2) * dv/dx
Here, v = (4x^2 + y^2)^2, so we have:
d/dx (tan^-4(4x^2 + y^2)^2) = -4 / (1 + ((4x^2 + y^2)^2)^2) * d/dx((4x^2 + y^2)^2)
3. Now we differentiate (4x^2 + y^2)^2:
Using the chain rule, we have d/dx ((4x^2 + y^2)^2) = 2(4x^2 + y^2) * d/dx(4x^2 + y^2)
Finally, we can combine all the derivatives together:
dy/dx = (6y^5 * 1 / (1 - (tan^-4(4x^2 + y^2)^2)^2) * (-4 / (1 + ((4x^2 + y^2)^2)^2) * 2(4x^2 + y^2)
Simplifying this expression further is not possible without specific values for x and y. However, this is the general expression for dy/dx based on the given equation.