How do you differentiate #y=sec^-1(x^7)#?

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Answer 1 · 2017-03-23T19:46:20

#(dy)/(dx)=7/(xsqrt(x^14-1))#

#y=sec^(-1) x^7#

#=>x^7=secy#

differentiate wrt #x#

#7x^6=secytany(dy)/(dx)#

#(dy)/(dx)=(7x^6)/(secytany#

we now substitute back to express the derivative as a function of #x#

#secy=x^7#

#1+tan^2y=sec^2y#

#tany=sqrt(sec^2y-1#

#(dy)/(dx)=(7x^6)/(x^7sqrt(sec^2y-1)#

#(dy)/(dx)=7/(xsqrt(x^14-1))#

Answer 2 · 2017-03-23T19:52:48

# dy/dx= 7/(xsqrt(x^14 -1)) #, or #7/(x^8sqrt(1-1/x^14)) #

Let

# y = sec^(-1)(x^7) #

Then:

# sec y = x^7#

Differentiating Implicitly:

# sec y tany dy/dx= 7x^6#
# :. x^7 tany dy/dx= 7x^6#
# :. tany dy/dx= 7x^6/x^7#
# :. tany dy/dx= 7/x#

And using the identity #sec^2 theta -= 1 + tan^2 theta # then;

# tan^2 y = sec^2y -1 #
# " " = (x^7)^2 -1 #
# " " = x^14 -1 #
# :. tan y =sqrt(x^14 -1) #

Substituting we get:

# sqrt(x^14 -1) dy/dx= 7/x #
# :. dy/dx= 7/(xsqrt(x^14 -1)) #

Which can also be written:

# dy/dx = 7/(xsqrt(x^14(1-1/x^14))) #
# " " = 7/(x*x^7*sqrt(1-1/x^14)) #
# " " = 7/(x^8sqrt(1-1/x^14)) #