Question

Using the concepts of Implicit Differentiation...
• find dy/dx of: "x + tan(xy) = 0"

I HAVE the final answer in my Calc textbook right in front of me, so I'm really not looking for someone to merely give that to me. I need the process, and the steps, explained and worked out that I may understand how these problem work- because let's face it, homework assignments have more than one problem, and I need to be able to do all of them. (If I wanted an easy fix, I could just copy answers on my own, true or not??)
How do you get to the final answer? On problems like these, where do you start? Any help is greatly appreciated! :)

*cough* (A medal and Fan status are up for grabs...) *cough*

Answer 1

@agent0smith Hey there buddy ole' pal... if you get a free moment during your school day today, would you mind taking a look at this? I'll be on OS later tonight for sure, if not sooner (relative to CST).

Answer 2

\[x+\tan xy=0\]
Differentiating, you have
\[1+\sec^2 (xy)\bigg[y+x\frac{dy}{dx}\bigg]=0\]
Derivative of \(x\) should be clear. The derivative of tangent is secant squared, but you also have that pesky \(xy\), an "inside" function that demands the use of the chain rule. The derivative of this function makes use of the product rule; \(\dfrac{d}{dx}x=1\) and \(\dfrac{d}{dx}y=\dfrac{dy}{dx}\).
\[\sec^2 (xy)\bigg[y+x\frac{dy}{dx}\bigg]=-1\\
y+x\frac{dy}{dx}=-\cos^2(xy)\\
x\frac{dy}{dx}=-\cos^2(xy)-y\\
\frac{dy}{dx}=-\frac{\cos^2(xy)+y}{x}\]