More derivatives (HL)

Here are some more derivatives for HL candidates.

function derivative function
tan⁡x\tan x sec⁡2x=1cos⁡2x\displaystyle \sec^2 x = \frac{1}{\cos^2x}
cosec⁡x\cosec x −cosec⁡xcot⁡x=−cos⁡xsin⁡2x\displaystyle -\cosec x\cot x = -\frac{\cos x}{\sin^2x}
sec⁡x\sec x sec⁡xtan⁡x=sin⁡xcos⁡2x\displaystyle \sec x\tan x = \frac{\sin x}{\cos^2x}
cot⁡x\cot x −cosec⁡2x=−1sin⁡2x\displaystyle -\cosec^2 x = -\frac{1}{\sin^2x}
ax=exln⁡aa^x = \e^{x\ln a} axln⁡aa^x\ln a
log⁡ax=ln⁡xln⁡a\displaystyle\log_ax = \frac{\ln x}{\ln a} \quad 1xln⁡a\displaystyle\frac {1}{x\ln a}
arcsin⁡x\arcsin x 11−x2\displaystyle \frac{1}{\sqrt{1-x^2}}
arccos⁡x\arccos x −11−x2\displaystyle -\frac{1}{\sqrt{1-x^2}}
arctan⁡x\arctan x 11+x2\displaystyle \frac{1}{1+x^2}

The derivatives of reciprocal trig functions automatically get an −- from from the −1-1 in f(x)−1f(x)^{-1}, but for sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x} the negative cancel with  d dxcos⁡x=−sin⁡x\frac{\d }{\d x}\cos x = -\sin x.

The derivatives of exponential and logarithmic functions applies chain rule to the function expressed in base ee.

The sign difference between derivatives of arcsin⁡x\arcsin x and arccos⁡y\arccos y speaks to the fact that the two inverse trig functions are mirror images of each other across horizontal line y=π4y = \frac{\pi}{4}. The inverse trig derivatives can be found using implicit differentiation and converting trig functions with unknown angle (SL).

Previous :
Next :