# 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?

• Try "derivative of x^2*y + y^5*Sec[x] = 5" at WolframAlpha. Commented Nov 21, 2012 at 18:14
• It ia fine. A bit too much detail. Two minor suggestions. Avoid $\frac{d}{dx}(y)$, use $\frac{dy}{dx}$ or $y'$. And avoid specially things like $\frac{d}{dx}(y)\sec x$, you do tat sort of thing several times. Inevitably this will be at some stage by you, or someone else, as the derivative of the product. Commented Nov 21, 2012 at 18:16
• I think I will choose $y'$ since it has the least amount of ambiguity, @AndréNicolas.
– 000
Commented Nov 21, 2012 at 18:25
• That worked perfectly. Thank you, @Amzoti.
– 000
Commented Nov 21, 2012 at 18:25
• Limitless, you are welcome. I would also recommend seeing if you can figure out how to transform the problem and arrive at some of the alternate forms that WA provided. Commented Nov 21, 2012 at 18:30

$$x^2y+y^5\sec(x)=5.$$ $$2xy+x^2y'+5y^4y'\sec(x)+y^5\sec'x=0$$ $$y'(x^2+5y^4\sec x)=-2xy-y^5\frac{\sin x}{\cos^2x }$$ $$y'=\frac{dy}{dx}=\frac{-2xy-y^5\frac{\sin x}{\cos^2x }}{x^2+5y^4\sec x}$$