Find
dy/dx
by implicit differentiation.
tan−1(4x2y) = x + 3xy2
Just use the product and chain rules:
arctan(4x^2y) = x + 3xy^2
1/(1+16x^4y^2) * (8xy+4x^2y') = 1 + 3y^2 + 6xyy'
8xy/(1+16x^4y^2) + 4x^2/(1+16x^4y^2) y' = 1+3y^2 + 6xyy'
y'(4x^2/(1+16x^4y^2) - 6xy) = 1+3y^2 - 8xy/(1+16x^4y^2)
At this point, it's just algebra, and serves no useful purpose to pursue it much further. Just solve for y' and you can massage it in various ways, as shown here:
http://www.wolframalpha.com/input/?i=derivative+arctan%284x^2y%29+%3D+x+%2B+3xy^2
Let's find dy/dx by differentiating both sides of the equation with respect to x:
d/dx(tan^(-1)(4x^2y)) = d/dx(x + 3xy^2)
Using the chain rule, the left side becomes:
1 / (1 + (4x^2y)^2) * d/dx(4x^2y)
Simplifying the left side:
1 / (1 + 16x^4y^2) * d/dx(4x^2y)
Using the product rule on the right side, we get:
1 / (1 + 16x^4y^2) * (8xy + 12x^2y^2 + 3y^2)
Combining like terms:
(8xy + 12x^2y^2 + 3y^2) / (1 + 16x^4y^2)
Therefore, dy/dx is equal to:
(8xy + 12x^2y^2 + 3y^2) / (1 + 16x^4y^2)
To find dy/dx by implicit differentiation, we will differentiate both sides of the equation with respect to x.
Starting with the given equation:
tan^(-1)(4x^2y) = x + 3xy^2
Differentiating both sides with respect to x:
d/dx [tan^(-1)(4x^2y)] = d/dx [x + 3xy^2]
To differentiate tan^(-1)(4x^2y), we use the chain rule. Let's calculate the left-hand side first:
Chain rule: d/dx [tan^(-1)(u)] = (1/ (1+u^2)) * d/dx [u]
Here, u = 4x^2y. Taking the derivative with respect to x:
d/dx [tan^(-1)(4x^2y)] = (1/ (1+(4x^2y)^2)) * d/dx [4x^2y]
Applying the chain rule again to d/dx [4x^2y]:
d/dx [4x^2y] = 4 * d/dx [x^2y] = 4 * (2xy + x^2(dy/dx))
Simplifying the expression:
(1/ (1+(4x^2y)^2)) * 4 * (2xy + x^2(dy/dx)) = 1 + 3y^2 + 3x * 2y(dy/dx)
Combining similar terms and rearranging the equation:
(1/ (1+16x^4y^2)) * (8xy + 4x^3y(dy/dx)) = 1 + 3y^2 + 6xy(dy/dx)
Now, let's solve for dy/dx by isolating the terms involving dy/dx:
(1/ (1+16x^4y^2)) * 4x^3y(dy/dx) - 6xy(dy/dx) = 1 + 3y^2 - 8xy
Factoring out dy/dx:
[(4x^3y - 6xy) / (1+16x^4y^2)] * dy/dx = 1 + 3y^2 - 8xy
Finally, solve for dy/dx by dividing both sides by the coefficient of dy/dx:
dy/dx = [(1 + 3y^2 - 8xy) / (4x^3y - 6xy)] * (1+16x^4y^2)
To find the derivative dy/dx by implicit differentiation, we will differentiate both sides of the equation with respect to x and then solve for dy/dx.
Step 1: Differentiate both sides of the equation with respect to x using the chain rule and product rule as needed.
Differentiating the left side:
d/dx (tan^(-1)(4x^2y)) = d/dx (x + 3xy^2)
Using the chain rule on the left side:
(1/(1 + (4x^2y)^2)) * d/dx (4x^2y) = 1 + 3y^2 + 6xy(dy/dx)
Simplifying the left side:
(1/(1 + 16x^4y^2)) * (8xy + 4x^2(dy/dx)y) = 1 + 3y^2 + 6xy(dy/dx)
Step 2: Simplify the equation and isolate dy/dx.
Multiplying both sides by (1 + 16x^4y^2) to eliminate the fraction on the left side:
8xy + 4x^2(dy/dx)y = (1 + 3y^2 + 6xy(dy/dx))(1 + 16x^4y^2)
Expanding the right side:
8xy + 4x^2(dy/dx)y = 1 + 3y^2 + 6xy(dy/dx) + 16x^4y^2 + 48x^5y^2(dy/dx)
Rearranging the equation to isolate dy/dx terms on one side:
8xy - 6xy(dy/dx) - 48x^5y^2(dy/dx) - 4x^2(dy/dx)y = 1 + 3y^2 - 16x^4y^2
Factoring out dy/dx terms:
dy/dx * (-6xy - 48x^5y^2 - 4x^2y) = 1 + 3y^2 - 16x^4y^2 - 8xy
Dividing both sides by (-6xy - 48x^5y^2 - 4x^2y) to solve for dy/dx:
dy/dx = (1 + 3y^2 - 16x^4y^2 - 8xy) / (-6xy - 48x^5y^2 - 4x^2y)
So, the derivative dy/dx by implicit differentiation is given by:
dy/dx = (1 + 3y^2 - 16x^4y^2 - 8xy) / (-6xy - 48x^5y^2 - 4x^2y)