Question

Question #2cc56

Answer 1

Use the product rule and the chain rule.

Explanation:

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)))`

Answer 2

Explanation:

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)`

Answer 3

Explanation:

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