Suppose that $x y ^ { 2 } + \sin y + x ^ { 3 } = 1$ .Find $\frac { d y } { d x }$ .
A) $- \frac { 1 } { 2 x y + \cos y }$
B) $- \frac { 3 x ^ { 2 } } { 2 x y + \cos y }$
C) $- \frac { 3 x ^ { 2 } + y ^ { 2 } } { 2 x y + \cos y }$
D) $- \frac { y ^ { 2 } } { 2 x y + \cos y }$

Question

Quizplus · Accepted Answer

To find $\frac{dy}{dx}$, we differentiate both sides of the equation $xy^2 + \sin y + x^3 = 1$ with respect to $x$, applying the chain rule and implicit differentiation. Differentiating $xy^2$ gives $y^2 + 2xy\frac{dy}{dx}$, differentiating $\sin y$ gives $\cos y \frac{dy}{dx}$, and differentiating $x^3$ gives $3x^2$. Setting the derivative of the right side (0, since the derivative of a constant is 0) equal to the sum of these derivatives gives $y^2 + 2xy\frac{dy}{dx} + \cos y \frac{dy}{dx} + 3x^2 = 0$. Solving for $\frac{dy}{dx}$ yields $\frac{dy}{dx} = -\frac{3x^2 + y^2}{2xy + \cos y}$.

Suppose That xy2+sin⁡y+x3=1x y ^ { 2 } + \sin y + x ^ { 3 } = 1xy2+siny+x3=1

Suppose That $x y ^ { 2 } + \sin y + x ^ { 3 } = 1$