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Find the equation of the tangent line: yx^2+xy^2=6, (1, 2). Can anyone explain the steps of implicit differentiation?
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y'x^2+y(2x)+x(2yy')+y^2=0
Thats the first step right?
y'(x^2+2xy)=-(2xy+y^2) y'=-y(2x+y)/x(x+2y)
m=y'(at (1,2))=-6/5
tangent line is y-2=-6/5(x-1)
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On the key it says the slope is -8/5
oh mistake in computations
-8/5 is the right ans
So implicit differentiation, can you quickly verbally go through it, what steps do you do?
for yx^2 use product rule (derivative of y)x^2+(derivative of x^2) y
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y'x^2+2xy....clear?
similar for the second term
OHHHHH!!!! I SEEEEEEE!!!!
THANK YOU SO MUCH! I really appreciate it :D
u welcome:)
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