How do you differentiate #y=csc^-1(4x^2)#?

1 Answer
Steve M
Nov 12, 2016

# dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1)))) #

Explanation:

# y=csc^-1(4x^2) <=> cscy=4x^2#

Differentiating implicitly we have:
# cscy=4x^2 #
# -cscycotydy/dx=8x #
# -(4x^2)cotydy/dx=8x #
# cotydy/dx=-2/x # ..... [1]

Now # 1 + cot^2A -= csc^2A #
# :. 1 + cot^2y = 16x^4 #
# :. cot^2y = 16x^4 - 1 #
# :. coty = sqrt(16x^4 - 1) #

Substituting into [1] we get:
# sqrt(16x^4 - 1)dy/dx=-2/x #

# :. dy/dx=(-2/x)/sqrt(16x^4 - 1) #
# :. dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1)))) #

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