Here are some derivatives you really need to be aware of for calc 2. Note:a >0 anda6= 1 andarcsin(x) = sin-1(x)ddxlnx=1xddxex=exddxlogax=1xlnaddxax=axlnaddxln[f(x)] =1f(x)·f0(x) =f0(x)f(x)ddxe[f(x)]=e[f(x)]·f0(x)ddxloga[f(x)] =1f(x) lna·f0(x) =f0(x)f(x) lnaddxa[f(x)]=a[f(x)]·f0(x) lnaddxsin-1(x) =1√1-x2ddxcos-1(x) =-1√1-x2ddxtan-1(x) =11 +x2ddxcot-1(x) =-11 +x2ddxsec-1(x) =1x√x2-1ddxsin-1(x) =-1x√x2-1I didn’t write them, but you should also be aware of the chain rule variants of the derivatives ofeach of the inverse trig functions so that, for example, you can find the derivative of sin-1(x5) orthe derivative of cot-1(x3·2x)You should also be aware that, depending on how sec-1(x) is defined, it’s derivative canalternatively be written asddxsec-1(x) =1|x|√x2-1. A similar fact is true for csc-1(x).For the inverse trig derivates an interesting observation can be made. Take for exampleddxsin-1(x) =1√1-x2. If we letx=sinθ, then the 1-x2in the derivative becomes1-x2= 1-sin2θ= cos2θ. That is, it simplifies nicely.The same is true for any of the other derivatives of the inverse trig functions, if we letxbe thecorresponding trig function ofθthen things simplify nicely.For sec-1(x), if we letx= secθthenx2-1 = sec2θ-1 = tan2θ. Try it for yourself with one ofthe other ones.
Important Inverse Trig derivatives you need to know: (notice that sometimes I use “arc” andsometimes “to the -1,” they both mean the same thing)ddxarcsin(x) =1√1-x2ddxarccos(x) =-1√1-x2ddxarctan(x) =11+x2ddxcot-1(x) =-11+x2ddxsec-1(x) =1|x|√x2-1ddxcsc-1(x) =-1|x|√x2-1(notice that the left side and right side derivatives only differ by a negaitve sign)Also if we want to write them for when we use the chain rule (hereuis a function ofx) we get:ddxarcsin(u) =1√1-u2·dudxddxarccos(u) =-1√1-u2·dudxddxarctan(u) =11+u2·dudxddxcot-1(u) =-11+u2·dudxddxsec-1(u) =1|u|√u2-1·dudxddxcsc-1(u) =-1|u|√u2-1·dudx