Question

use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point

Answer 1

\[\arctan(xy)=\arcsin(x+y), (0,0)\]

Answer 2

\[\frac{dy}{dx}=\frac{x^2 y^2-y \sqrt{1-(x+y)^2}+1}{-x^2 y^2+x \sqrt{1-(x+y)^2}-1} \]The RHS above evaluated at point (0,0) is -1.

Answer 3

could you explain how you got that? I really am confused on this problem

Answer 4

Used Mathematica for the calculations.
Took the total derivative of the problem equation.\[\frac{y dx+x dy}{x^2 y^2+1}=\frac{dx+dy}{\sqrt{1-(x+y)^2}} \]Solved for dy.\[dy=\frac{\frac{dx}{\sqrt{1-(x+y)^2}}-\frac{y dx}{x^2 y^2+1}}{\frac{x}{x^2 y^2+1}-\frac{1}{\sqrt{1-(x+y)^2}}} \]Factored dx out of the RHS numerator and then divided both sides of the equation by dx.\[\frac{dy}{dx}=-\frac{-x^2 y^2+y \sqrt{1-(x+y)^2}-1}{\left(x^2 y^2+1\right) \sqrt{1-(x+y)^2} \left(\frac{x}{x^2 y^2+1}-\frac{1}{\sqrt{1-(x+y)^2}}\right)} \]Simplified the RHS.\[\frac{dy}{dx}=\frac{x^2 y^2-y \sqrt{1-(x+y)^2}+1}{-x^2 y^2+x \sqrt{1-(x+y)^2}-1} \]Evaluated the above at (0,0).\[\frac{dy}{dx}=\frac{0+1}{0-1}=-1 \]

Answer 5

I dont doubt what you are saying, but I haven't been taught what you did above. we are learning how to take derivatives of inverse trig functions, and I am a bit confused. what is RHS. also it says to do without a calculator. the example they refer to that is related to this problem in the text book shows them taking the tangent of both sides to eliminate one , then differentiating the other side using applicable rules eg. product rule quotient rule etc. On another they use a triangle and Pythagorean theorem

Answer 6

RHS is short hand for "Right Hand Side", referring to the right hand side of an equation.

Total Derivative:
http://www.youtube.com/watch?v=7xhAzlJKxY8