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the slope of the tangent line is given by the dy/dx. So this question is on implicit differentiation
yea i'm only having problems differentiatiog the right side of the equation
2(x^2+y^2)^2=25(x^2-y^2) Differentiate both sides with respect to x: 4(x^2 + y^2)[2x + 2y(dy/dx)] = 25[2x - 2y(dy/dx)]
Do you know how to continue from there?
uhmm not quite
but for the right side isn't it 25 (2x-y^2 y' +x^2-2y)
The slope of the tangent line is given by the dy/dx, and now the question is to determine the slope at (-3,1). So substitute x= -3 and y = 1 into the equation and then rearrange it to get the dy/dx. Would you like to try it first?
Why is it 25 (2x-y^2 y' +x^2-2y)?
multiplying 25 by the derivative of x^2-y^2+x^2- the derivative of y^2
the right side is only 25(x^2 - y^2). Differentiate it you have (d/dx)[25(x^2 - y^2)] = 25 (d/dx)[x^2 - y^2] = 25 [(d/dx)(x^2) - (d/dx)(y^2)] = 25 [2x - 2y (dy/dx)]
Sorry, i still don't understand how did you get yours...
So are you able to continue it now? :)
still stuck :(
ok, we have x=-3 and y=1, so substituting in we get 4(x^2 + y^2)[2x + 2y(dy/dx)] = 25[2x - 2y(dy/dx)] 4(9 + 1)[(-6) + 2(dy/dx)] = 25[(-6) - 2(dy/dx)] -240 + 80(dy/dx) = -150 - 50(dy/dx) 130 (dy/dx) = 90 dy/dx = 9/13
oh i thought i had to simplify until i find y' then i substitute in... oh well it makes sense though, thanks :)
you can also rearrange to find your y' then only you substitute. But I think in this case, it is easier to substitute to get your y'