find y' if arctan(xy)=6+(x^4)y
To find y', we need to take the derivative of both sides of the equation with respect to x. Let's start with the given equation:
arctan(xy) = 6 + (x^4)y
To differentiate arctan(xy) with respect to x, we will apply the chain rule. The derivative of arctan(u) with respect to u is 1/(1 + u^2), and u = xy in this case. So, we have:
d/dx(arctan(xy)) = d/d(xy) * xy * 1/(1 + (xy)^2)
Now, let's differentiate the right side of the equation with respect to x. We have two terms: 6 and (x^4)y. The derivative of 6 with respect to x is 0 since it is a constant. To differentiate the second term, we use the product rule:
d/dx((x^4)y) = (d/dx(x^4)) * y + (d/dx(y)) * x^4
Differentiating x^4 gives us:
d/dx(x^4) = 4x^3
Since we are finding y', we want to isolate dy/dx (the derivative of y with respect to x). Rearranging the equation, we have:
d/d(xy) * xy * 1/(1 + (xy)^2) = 4x^3 * y + (d/dx(y)) * x^4
Now, let's solve for (d/dx(y)) by isolating it on one side:
(d/dx(y)) * x^4 = d/d(xy) * xy * 1/(1 + (xy)^2) - 4x^3 * y
Finally, divide both sides by x^4 to solve for (d/dx(y)):
d/dx(y) = [d/d(xy) * xy * 1/(1 + (xy)^2) - 4x^3 * y] / x^4
This expression represents the derivative of y with respect to x, denoted as y'.