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

2 Answers
sjc
Mar 23, 2017

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

Explanation:

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))

Steve M
Mar 23, 2017

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

Explanation:

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))

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