<p>Do you guys know the partial derivative trick for doing implicit differentiation insanely fast? </p>

<p>say you have an equation like tan(y)-sin(e^(y*x))=1 and you want to find dy/dx.</p>

<p>move everything to the side so you get an expression for the form F(x,y)=0. In this case: </p>

<p>tan(y)-sin(e^(y<em>x))-1=0 (</em>)</p>

<p>Now differentiate this equation in respect to x while pretending y is a constant (this is called partial differentiation...a topic normally taught in multivariable calculus).</p>

<p>you get -y<em>e^(x</em>y)<em>cos(e^(x</em>y)). (1)</p>

<p>Now do the same to (*) except with respect to y: that is, differentiate with respect to y while pretending x is a constant:</p>

<p>(secy)^2-x<em>e^(x</em>y)<em>cos(e^(x</em>y)). (2)</p>

<p>Now, the trick is this: multiply (1) by -1 and divide it by (2). You get:</p>

<p>dy/dx= [-y<em>e^(x</em>y)<em>cos(e^(x</em>y)<em>(cosy)^2]/[x</em>e^(x<em>y)</em>cos(e^(x<em>y)</em>(cosy)^2-1]</p>

<p>Tada! a seemingly impossible implicit differentiation question can be really easy using this trick. Here's a generalization</p>

<p>If you have F(x,y)=0 and you want dy/dx, </p>

<p>dy/dx=-F<em>x/F</em>y where F_t denotes the partial derivative of F with respect to t. </p>

<p>Skeptical? Try this one. Given x^2+y^2=1, find dy/dx.</p>

<p>F(x,y):=x^2+y^2-1. now, pretending y is a constant, we have F<em>x=2x

pretending x is a constant, we have F</em>y=2y. </p>

<p>now, dy/dx=-(2x)/(2y)=-x/y. If you do the whole implicit differentiation thing, you'll see you end up with the same answer.</p>

<p>Cool, no?</p>