Solve for $\frac{dy}{dx}$?

Question

Q: Given $3^{x+y} = x^3 + 3y$, find $\frac{dy}{dx}$.

I am convinced that, since it is not possible to algebraically solve for $y(x)$, one can't find $\frac{dy}{dx}$.

Am I correct?

Thanks!

Answer 1

Let$$f=3^{x+y}-x^{3}-3y=0$$ $$\frac{dy}{dx}=\frac{-\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} $$ $$\frac{dy}{dx}=\frac{-3^{x+y}ln3+3x^{2}}{3^{x+y}ln3-3}$$

I think so this might be enough! Please correct me if I am wrong.

Answer 2

Using the chain rule and implicit differentiation, we can write

\begin{equation*} 3^{x+y}\log(3)(\frac{d}{dx}(x+y))=\frac{d}{dx}(x^3+3y)\\ \Rightarrow \frac{d}{dx}(x)+\frac{d}{dx}(y)3^{x+y}\log(3)=\frac{d}{dx}(x^3+3y)\\ \Rightarrow 3^{x+y}\log(3)(1+y'(x))=\frac{d}{dx}(x^3+3y)\\ \Rightarrow 3^{x+y}\log(3)(1+y'(x))=3(\frac{d}{dx}(y))+3x^2\\ \Rightarrow 3^{x+y}\log(3)+3^{x+y}\log(3)y'(x)=3x^2+3y'(x). \end{equation*}

Doing some manipulation and rearranging gives us

\begin{equation*} y'(x)=\frac{3x^2-3^{x+y}\log(3)}{-3+3^{x+y}\log(3)}. \end{equation*}