We have two responses for you
Let me try a similar expression,
(x2 + y2)2 = (x - y)2
I want to find the slope of the tangent to this curve at (1, -1).
First I should check that the point is on the curve. At (1, -1) the left side is (12 + (-1)2)2 = 22 = 4 and the right side is (1 - (-1))2 = 22 = 4. Thus (1, -1) is on the curve.
The slope of the tangent to the curve at (1, -1) is the derivative of y with respect to x, evaluated at (1, -1). When I differentiate I need to remember that y is a function of x.
Differentiate the left side with respect to x.
Using the chain rule d/dx[(x2 + y2)2] = 2(x2 + y2) d/dx(x2 + y2)
I now need to differentiate (x2 + y2) and I need to remember to use the chain rule again when I get to d/x(y2). Thus d/x(y2) = 2 y dy/dx. Hence putting it all together I get
d/dx[(x2 + y2)2] = 2(x2 + y2) d/dx(x2 + y2) = 2(x2 + y2) (2x + 2 y dy/dx)
Differentiate the right side with respect to x.
d/dx[(x - y)2] = 2(x - y) d/dx(x - y) = 2(x - y) (1 - dy/dx)
Since the left side is equal to the right side
2(x2 + y2) (2x + 2 y dy/dx) = 2(x - y) (1 - dy/dx)
To complete the problem I need to find dy/dx at (1, -1) so I substitute x = 1, y = -1 into the equation above to get
2(1 + 1) (2 - 2 dy/dx) = 2(1 + 1) (1 - dy/dx)
2(2)(2 - 2 dy/dx) = 4(1 - dy/dx)
divide both sides by 4 to get
2 - 2 dy/dx = 1 - dy/dx
dy/dx = 1.
Now try your expression and write back if you still have difficulties.
- I'm assuming the "2" just before the equals sign is
an exponent, and have made the correction above.
To solve a problems like this, use implicit differentiation.
Let me illustrate with a simple example:
Find the slope of x4 + y6 = 1 + y2 at (1,1)
Take the derivative of each side with respect to x. When you hit a y, use the chain rule and leave the y' in place:
4x3 + 6 y5 y' = 2 y y'
Now isolate all the terms with y' on one side and factor out the y':
4 x3 = (2y - 6y5) y'
divide to isolate y' and plug in the coordinates to evaluate:
y' = 4 x3 /(2y - 6y5) = -1
- Implicit differentiation is also a very powerful technique in solving related rates problems and constrained optimization problems (It's not used as widely as it should be; too many students are nervous of it and try to eliminate a variable. Make it your friend - you'll be glad you did.) In applications, it's very helpful in thermodynamics, theoretical mechanics, and economics.