find y' if arctan(xy)=6+x^(4)y
here is one way:
arctan(xy)=6+x^(4)y
or
xy = tan(6 + x^4y)
xdy/dx + y = sec^2 (6 + x^4y) (x^4 dy/dx + 4x^3y_
= x^4 dy/dx sec^2 (6+x^4y) + 4x^3y sec^2(6+x^4y)
dy/dx(x - x^4sec^2 (6+x^4y) ) = 4x^3 sec^2 (6+x^4y) - y
dy/dx = (4x^3 sec^2 (6+x^4y) - y)/(x - x^4sec^2 (6+x^4y) )
check my algebra, I should have written it out on paper first.
Or, more directly,
1/(1+x^2y^2) (y+xy') = 4x^3y+x^4y'
y+xy' = 4x^3y (1+x^2y^2) + x^4(1+x^2y^2)y'
y' = (4x^3y + 4x^5y^3-y)/(x-x^4-x^6y^2)
This can also be obtained (after restoring the y that was accidentally dropped) from Reiny's result by recalling the sec^2 = 1+tan^2 = 1+x^2y^2
What do mean
To find y', we need to differentiate both sides of the equation with respect to x using the rules of differentiation. Let's start with the equation:
arctan(xy) = 6 + x^4 * y
In order to differentiate, it will be helpful to rewrite the equation in a different form. Let's separate the terms involving x and y:
arctan(xy) - x^4 * y = 6
Now, we can differentiate both sides of the equation with respect to x:
Differentiating the left-hand side:
To differentiate arctan(xy), we can use the chain rule. Let's define f(u) = arctan(u), and u = xy. Then,
f'(u) = 1 / (1 + u^2) (derivative of arctan(u))
Now, applying the chain rule:
(d/dx) [arctan(xy)] = f'(u) * (d/dx) [xy]
= f'(xy) * (d/dx) [xy]
= f'(xy) * (xdy/dx + y)
Differentiating the right-hand side:
The derivative of a constant (6) with respect to x is zero.
The derivative of x^4 * y with respect to x can be found using the product rule. Let's define f(x) = x^4 and g(x) = y, then:
(d/dx) [x^4 * y] = f'(x) * g(x) + f(x) * g'(x)
= 4x^3 * y + x^4 * (dy/dx)
Now, let's put it all together and solve for y':
f'(xy) * (xdy/dx + y) - x^4 * (dy/dx) - 4x^3 * y = 0
Let's rearrange the terms to solve for dy/dx (y'):
f'(xy) * x * dy/dx - x^4 * dy/dx - 4x^3 * y = -f'(xy) * y
Now, factor out dy/dx (y'):
(dy/dx) * [f'(xy) * x - x^4] = -f'(xy) * y + 4x^3 * y
Finally, divide both sides by [f'(xy) * x - x^4]:
dy/dx = [-f'(xy) * y + 4x^3 * y] / [f'(xy) * x - x^4]
Now, substitute f'(xy) = 1 / (1 + (xy)^2):
dy/dx = [-y / (1 + (xy)^2) + 4x^3 * y] / [x / (1 + (xy)^2) - x^4]
Simplifying the expression, we get the derivative dy/dx (y'):
dy/dx = [4x^3 * y - y] / [x - x^4(1 + (xy)^2)]