What is the derivative of ( tanx+cotx)/(tanx-cotx)Answer is: #-2sec2x tan2x#. How to get this answer??Please help me
What is the derivative of ( tanx+cotx)/(tanx-cotx)Answer is: -2sec2x tan2How to get this answer.Please help me
What is the derivative of ( tanx+cotx)/(tanx-cotx)Answer is: -2sec2x tan2How to get this answer.Please help me
2 Answers
Explanation:
Let's Apply the Quotient Rule here.
[The Quotient Rule States That if
Let's assume
So,
Finally Got The Answer Right. Thanks Everyone Who helped me!
And, Hope this Helps.
# d/dx (tanx+cotx)/(tanx-cotx) = -2sec2xtan2x #
Explanation:
For convenience, define:
# y = (tanx+cotx)/(tanx-cotx) #
Before we differentiating we can simplify the expression:L
# y = (sinx/cosx+cosx/sinx)/(sinx/cosx-cosx/sinx) #
# \ \ = ((sin^2x+cos^2x)/(sinxcosx))/((sin^2x-cos^2x)/(sinxcosx)) #
# \ \ = ((sin^2x+cos^2x)/(sinxcosx))/((sin^2x-cos^2x)/(sinxcosx)) #
# \ \ = (sin^2x+cos^2x)/(sinxcosx) * (sinxcosx)/(sin^2x-cos^2x) #
# \ \ = (sin^2x+cos^2x)/(sin^2x-cos^2x) #
And we can use two fundamental trigonometric identities:
# sin^2x+cos^2x -= 1 \ \ \ \ \ \ \ \ \ \# (Pythagorean Identity)
# cos2x -= cos^2x - sin^2x \ \ # (Multiple Angle Identity)
Giving us
# y = (1)/(-cos2x) = -sec2x #
And now we differentiate wrt
# dy/dx = (-sec2xtan2x)(2)#
Thus:
# d/dx (tanx+cotx)/(tanx-cotx) = -2sec2xtan2x \ \ \ # QED