How do you differentiate g(x) =4/cosx + 1/ tanx g(x)=4cosx+1tanx?
1 Answer
Jan 31, 2016
Explanation:
Assuming you don't know the identities for the derivatives of secant and cotangent:
We can find these derivatives using the quotient rule.
g'(x)=(cosxd/dx(4)-4d/dx(cosx))/cos^2x+(tanxd/dx(1)-1d/dx(tanx))/tan^2x
Recall that
g'(x)=(4sinx)/cos^2x+(-sec^2x)/tan^2x
g'(x)=4(sinx/cosx)1/cosx-1/cos^2x(cos^2x/sin^2x)
g'(x)=4tanxsecx-csc^2x
Assuming you already know the identities:
g(x)=4secx+cotx
Thus,
g'(x)=4secxtanx-csc^2x
since
d/dx(secx)=secxtanx andd/dx(cotx)=-csc^2x