Question

If $x^3{y}^3+3x\sin y=e^y$, then $\frac{dy}{dx}=?$

• A. $\frac{3x^2{y}^3+3\sin y}{e^y-3x^3{y}^2-3x\cos y}$
• B. $\frac{3x^2{y}^3+3\sin y}{e^y+3x^3{y}^2+3x\cos y}$
• C. $\frac{3x^2{y}^3-3\sin y}{e^y+3x^3{y}^2-3x\cos y}$
• D. $\frac{3x{y}^3+3\cos y}{e^y-3x^3{y}^2-3x\cos y}$

Answer 1

The correct option is A $\frac{3x^2{y}^3+3\sin y}{e^y-3x^3{y}^2-3x\cos y}$

$3x^2{y}^3+3x^3{y}^2\frac{dy}{dx}+3\sin y+3x\cos y\frac{dy}{dx}=e^y\frac{dy}{dx}$

$3x^2{y}^3+3\sin y=(e^y-3x^3{y}^2-3x\cos y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{3x^2{y}^3+3\sin y}{e^y-3x^3{y}^2-3x\cos y}$