Question #2cc56

Given: #f(x)= x sec^-1 (x^3)#

The product rule is:

#d/dx(u*v) = (du)/dx*v+ u*(dv)/dx#

For the given function:

Let #u = x# and #v = sec^-1(x^3)#

#(du)/dx = 1#

The computation of #(dv)/dx# requires the use of the chain rule

#(d(g(h(x))))/dx = (dg)/(dh)(dh)/dx#

Let #h(x) = x^3# and #g = sec^-1(h)#

#(dh)/dx = 3x^2#

#(dg)/(dh) = 1/(h^2sqrt(1 - 1/h^2)#

Substitute these into the right side of the chain rule:

#(d(g(h(x))))/dx = (1/(h^2sqrt(1 - 1/h^2)))(3x^2)#

Reverse the substitution for h:

#(d(g(h(x))))/dx = (1/((x^3)^2sqrt(1 - 1/(x^3)^2)))(3x^2)#

Two powers of x cancel and the 3 can move to the numerator:

#(d(g(h(x))))/dx = 3/((x^4)sqrt(1 - 1/(x^6)))#

Returning to the product rule:

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

Substitute the values into the product rule:

#d/dx(xsec^-1(x^3)) = sec^-1(x^3) + x*(3/((x^4)sqrt(1 - 1/(x^6))))#

one power of x cancels in the second term:

#d/dx(xsec^-1(x^3)) = sec^-1(x^3) + 3/((x^3)sqrt(1 - 1/(x^6)))#

We know that differential of inverse trigonometric ratio
#g(x)=sec^-1(x)#
#g'(x) = 1 / (x sqrt(x ^2 - 1))#

#:.# if #g(x)=sec^-1 (x^3) #
We get #g'(x) = 1 / (x^3 sqrt((x^3) ^2 - 1))# ........(1)

Given expression is
#f(x)= x sec^-1 (x^3)#
Using chain rule we get
#f'(x)= x d/dx(sec^-1 (x^3))+(sec^-1 (x^3))xxd/dx x#
Using (1) we get
#f'(x)= x xx(3x^2) xx(1 / (x^3 sqrt((x^3) ^2 - 1)))+(sec^-1 (x^3))xx1#
#=>f'(x)= (3x^3)/ (x^3 sqrt(x^6 - 1))+sec^-1 (x^3)#
#=>f'(x)= 3/ ( sqrt(x^6 - 1))+sec^-1 (x^3)#

make use of this to write for #tan(y/x)=tan(sec^-1(x^3))=sqrt(x^6-1)#