How do you differentiate g(x) =4/cosx + 1/ tanx g(x)=4cosx+1tanx?

1 Answer
Jan 31, 2016

g'(x)=4tanxsecx-csc^2x

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 d/dx("constant")=0, d/dx(cosx)=-sinx and d/dx(tanx)=sec^2x.

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) can be rewritten as

g(x)=4secx+cotx

Thus,

g'(x)=4secxtanx-csc^2x

since

d/dx(secx)=secxtanx and d/dx(cotx)=-csc^2x