(-x+tanx)'=-1+(1/cos2x)
Chat with our AI personalities
d/dx(1+tanx)=0+sec2x=sec2x
(tan x- 1)/ (1+tan x)
y' = (sec(x))^2
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x