Use implicit differentiation to find the slope (m) of the tangent line to the curve
y/(x+6y)=x^2-5
at the point (1, -4/25)
by 
Anonymous
ANSWERS: 1
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y/(x+6y)=x^2-5 USING THE QUOTIENT RULE. [(y')(x+6y)-(y)(1+6(y'))/(x+6y)^2]=2x (y')(x+6y)-(y)(1+6(y')) =2x(x+6y)^2 (y')(x+6y)-y-6y(y')) =2x(x+6y)^2 (y')(x+6y)-6y(y')) =2x(x+6y)^2+y (y')[x+6y-6y] =2x(x+6y)^2+y y'= dy/dx = ={[2x(x+6y)^2]+y}/[x+6y-6y] now plug in (1, -4/25) for (x,y) and you got your slop. (= (dy/dx))
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