For y3=x2+7y^3 = x^2 + 7y3=x2+7, dy/dxdy/dxdy/dx at the point (1,2)(1, 2)(1,2) isA23\tfrac{2}{3}32B13\tfrac{1}{3}31C16\tfrac{1}{6}61check_circleD222Explanation3y2y′=2x⇒y′=2x/(3y2)3y^2 y' = 2x \Rightarrow y' = 2x/(3y^2)3y2y′=2x⇒y′=2x/(3y2). At (1,2)(1, 2)(1,2): 2/(3⋅4)=1/62/(3 \cdot 4) = 1/62/(3⋅4)=1/6.