Inverse Trig Derivatives

<p>Hi,</p>

<p>I completely memorized derivatives of all the basic trigs (sin, cos, tan, csc, sec, cot) and every non-trig derivative like ln(x), e^x, etc.; I have yet to memorize derivatives of inverse trig functions.</p>

<p>I don't want to spend time trying to memorize all the ugly inverse trig derivatives, but will not knowing them cost you some points? I'm guessing that the Exam will mostly test you on basic derivatives like I mentioned above, but I'm worried.
Would it be all right to go into the Exam without knowing the inverse trig derivatives?</p>

<p>Acutally, inverse trig derivatives are included as 'basic' derivatives on the exam. You'll need to be familiar with them.</p>

<p>What are the best ways of memorizing the inverse trig derivatives?</p>

<p>The inverse trig derivatives are not nearly as important as the basic 6, or even 3. You might see a question or two on it on multiple choice, and my teacher said that the college board isn't evil enough to make it into a free-response. (Knock on wood)</p>

<p>It's good to know them, but there are more important things to learn.</p>

<p>It's really not that hard to learn them, just memorize them like everything else.</p>

<p>Try deriving them yourself using trig substitution and such, it'll help you remember it better.</p>

<p>if you know sine and cosine, you can derive the rest. same with all trig stuff.</p>

<p>I still think they're easier to memorize. They can come in very handy for no-calc integration problems. For example, if you are asked to integrate 1/(x^2+1), which happens to be inverse tangent.</p>

<p>hmmm
sin is the FIRST trig function, so the 1 comes FIRST in the derivative of inverse sin (and you subtract x^2 b/c if it was addition no one would care which came first)</p>

<p>I remember that my favorite is tan, so it has no sqrt and is addition, nice and easy.</p>

<p>and , secant is sneaky (with the absolute value) and it is like the on the "opposite" trig functions. So it is opposite to sin (actualy cos i guess) b/c i has addition.</p>

<p>cofunctions are just the neg of their originals...</p>