What is the derivative of this function #arcsec(x^3)#?

Question

Scarlett Allison · Accepted Answer

#=(3x)/sqrt(x^2-1)#

For this you need to know what the derivative of arcsec(x) is.

You can derive it fairly easily:

#y = arcsec(x)#
#sec(y) = x#
#d/dx sec(y) = d/dx x#
#sec(y)tan(y)dy/dx = 1#
#dy/dx = 1/(sec(y)tan(y)#
#dy/dx = 1/(sec(arcsec(x))tan(arcsec(x)))#
#dy/dx = 1/(xsqrt(x^2-1))#

In the last statement I simplified #sec(arcsec(x))# to #x#, which is obvious, but I also simplified #tan(arcsec(x))# to #sqrt(x^2-1)# which is slightly less obvious

If you draw a triangle such that sec(angle) = x, you can understand why it works:

you can see that #arcsec(x) = theta# and #tan(theta) = sqrt(x^2-1)#

#therefore tan(arcsec(x)) = sqrt(x^2-1)#

Now we know the derivative of arcsec(x) and just need to apply a little chainrule to get our answer:

#d/dx arcsec(x^3) = 1/(xsqrt(x^2-1))*d/dx x^3 = (3x^cancel2)/(cancelxsqrt(x^2-1))#
#=(3x)/sqrt(x^2-1)#

Emma Aldridge · Answer

#dy/dx = 3/(absxsqrt(x^6-1))#

If #y=arcsec(x^3)# then #secy =x^3#, that is: #cosy=1/x^3#

Differentiate implicitly:

#-sinydy/dx = -3/x^4#

#dy/dx = 3/(x^4siny)#

Now, for #0 <= y <= pi #:

#siny = sqrt(1-cos^2y) = sqrt(1-1/x^6) #

and:

#dy/dx = 3/(x^4 sqrt(1-1/x^6)) = (3sqrt(x^6))/(x^4sqrt(x^6-1))= 3/(absxsqrt(x^6-1))#

Cameron Alexander · Answer

undefined The derivative of the function $ \text{arcsec}(x^3) $ with respect to $ x $ is $ \frac{3x^2}{|x^6|\sqrt{x^6-1}} $.