How do you find the derivative of # y = tanx + cotx #?

2 Answers
Feb 20, 2017

#f'(x)=sec^2(x)-csc^2(x)#

Explanation:

Treat the tanx separately and the cotx separately. Don't forget the derivatives for trig functions:
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Derivative of #tanx# is #sec^2x#. Derivative of #cotx# is #-csc^2x#. Since they are adding together, we can treat them it as #tan'x# and #cot'(x)#:

#f'(x)=sec^2(x)-csc^2(x)#

Feb 20, 2017

#y' = -cos2xsec^2xcsc^2x#

Explanation:

A bit of a different approach...

Rewrite in terms of sine and cosine.

#y = sinx/cosx + cosx/sinx#

#y = (sin^2x + cos^2x)/(sinxcosx)#

#y = 1/(sinxcosx)#

#y = (sinxcosx)^-1#

Now use the chain rule to differentiate. Let #y = u^-1# and #u = sinxcosx#. The function #u# can be differentiated using the product rule to #(du)/dx = cosx(cosx) + sinx(-sinx) = cos^2x - sin^2x = cos2x#. By the power rule, #dy/(du) = -1/u^2#.

#y' = dy/(du) * (du)/dx#

#y' = cos2x * -1/u^2#

#y' = -cos(2x)/(sinxcosx)^2#

#y' = -cos2xsec^2xcsc^2x#

Hopefully this helps!