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

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Answer 1 · 2016-01-31T15:55:24

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

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#