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find dy/dx xy^2=16
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Let's use implicit differentiation: x(2y)(y') + y^2 = 16 2xy(y') = 16 - y^2 dy/dx = (16 - y^2)/2xy
shouldnt it be -y^2/2xy
Sorry yes.
its okay and thank you :D
By the way you may simplify to -y/2x
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y*-y/2*x*y
i was just bout to write that
find dy/dx of y(x+y)=4
\[ yx+y^2=0\] \[dy/dx1*x+y+dy/dx2y=0\] \[-y/x+2y\]
\[\frac{dy}{dx}=\frac{y}{x+2y}\]
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