Derivatives Impicit

Question

(1) Use implicit differentiation to find the derivative of : (1) xy^3/1+y = e^xy I got my final answer as dy/dx = [y^6 +xy^3 - ye^xy ( y+1)^2] / [ ( xe^xy) (y+1)^2 - 3xy^5] Anyone got another answer with correct steps?

Answer 1

Our starting point is $$ \frac{xy^3}{1 + y} = e^{xy}$$ Differentiating both sides w.r.t. $x$, $$\frac{d}{dx} \left( \frac{xy^3}{1 + y} \right) = \frac{d}{dx} \left( e^{xy} \right)$$ $$ \Rightarrow \frac{y^3}{1 + y} + x \cdot \frac{dy}{dx} \frac{d}{dy} \frac{y^3}{1 + y} = \frac{d(xy)}{dx} \cdot \frac{d}{d(xy)} e^{xy}$$ $$\Rightarrow \frac{y^3}{1 + y} + x \cdot \frac{dy}{dx} \frac{3y^2(1 + y) - y^3}{(1 + y)^2} = \left(y + x \frac{dy}{dx} \right) \cdot e^{xy}$$ $$\Rightarrow \frac{y^3}{1 + y} + x \cdot \frac{dy}{dx} \frac{2y^3 + 3y^2}{(1 + y)^2} = \left(y + x \frac{dy}{dx} \right) \cdot e^{xy}$$ $$\Rightarrow \frac{dy}{dx} \left( x \cdot \frac{2y^3 + 3y^2}{(1 + y)^2} - x \cdot e^{xy} \right) = y \cdot e^{xy} - \frac{y^3}{1 + y}$$ Now from here there are many variations of the final answer depending on how you want to use the substitution $ \frac{xy^3}{1 + y} = e^{xy}$. For example, we can write $$\frac{2xy^3}{(1 + y)^2} = \frac{xy^3}{1 + y} \cdot \frac{2}{1 + y} = \frac{2e^{xy}}{1 + y} $$ So it's a bit hard to verify the answer that you arrived at. Perhaps you could compare with the above and let us know if you're more confident in your answer now.

Answer 2

When I face this kind of problem, I personally prefer to use the implicit function. In your case $$F=\frac{xy^3}{1 + y} - e^{xy}=0$$ Now, compute each derivative $$F'_x=\frac{y^3}{1 + y} - ye^{xy}$$ $$F'_y=\frac{x y^2 (2 y+3)}{(y+1)^2}-x e^{x y}$$ Now, using the implicit function theorem $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{y (y+1) \left((y+1) e^{x y}-y^2\right)}{x y^2 (2 y+3)-x (y+1)^2 e^{x y}}$$