<s>[INST] Implicit Differentiation of $x^2y+y^5\sec(x)=5$. Problem
I have began teaching myself Calc I and I've come across the following problem:

Find $\frac{dy}{dx}$ for the following:
  $$x^2y+y^5\sec(x)=5.$$

I automatically presumed this was an implicit differentiation. However, since I'm somewhat new to implicit differentiation, my solution looks messy. It is below, and my question is: Am I doing this right and are there ways I could improve (whether it be in terms of my notation, method, or something else)?
Solution
\begin{align}
x^2y+y^5\sec(x)&=5\\
\frac{d}{dx}\left(x^2y+y^5\sec(x)\right)&=\frac{d}{dx}5\\
\frac{d}{dx}x^2y+\frac{d}{dx}y^5\sec(x)&=\frac{d}{dx}5\\
2xy+x^2\frac{d}{dx}(y)+\frac{d}{dx}(y^5)\sec(x)+y^5\sec(x)\tan(x)&=0\\
2xy+x^2\frac{d}{dx}(y)+5y^4\frac{d}{dx}(y)\sec(x)+y^5\sec(x)\tan(x)&=0\\
x^2\frac{d}{dx}(y)+5y^4\frac{d}{dx}(y)\sec(x)&=-2xy-y^5\sec(x)\tan(x)\\
\frac{d}{dx}(y)\left(x^2+5y^4\sec(x)\right)&=-2xy-y^5\sec(x)\tan(x)\\
\frac{dy}{dx}&=\frac{-2xy-y^5\sec(x)\tan(x)}{x^2+5y^4\sec(x)}.
\end{align}
Side question
How could I put this into Wolfram Alpha?
 [/INST] $$x^2y+y^5\sec(x)=5.$$
$$2xy+x^2y'+5y^4y'\sec(x)+y^5\sec'x=0$$
$$y'(x^2+5y^4\sec x)=-2xy-y^5\frac{\sin x}{\cos^2x }$$
$$y'=\frac{dy}{dx}=\frac{-2xy-y^5\frac{\sin x}{\cos^2x }}{x^2+5y^4\sec x}$$
</s>