Question

Find the first derivative of: y = x^4 sin^-1 (x^4)

Answer 1

Given:

`y=x^4\sin ^(-1)(x^4)`

Differentiate with respect to x.

`(dy)/(dx)=(d(x^4\sin ^(-1)(x^4)))/(dx)`

`\text{Use (uv)'=uv'+vu', here u=x}^4\text{ and v=}\sin ^(-1)(x^4).`

`(dy)/(dx)=\sin ^(-1)(x^4)(d(x^4))/(dx)+x^4(d(\sin^(-1)(x^4)))/(dx)`
`\text{ Use }(dx^4)/(dx)=4x^3\text{ and }(d(\sin^(-1)(x^4)))/(dx)=\frac{(dx^4)/(dx)}{\sqrt[]{1-(x^4)^2}}\text{.}`

`(dy)/(dx)=\sin ^(-1)(x^4)*4x^3+x^4\frac{(dx^4)/(dx)}{\sqrt[]{1-(x^4)^2}}`

`(dy)/(dx)=\sin ^(-1)(x^4)*4x^3+x^4\frac{4x^3}{\sqrt[]{1-x^8}}`

`(dy)/(dx)=4x^3\lbrack\sin ^(-1)(x^4)+\frac{x^4}{\sqrt[]{1-x^8}}\rbrack`

Hence the first derivative of the given equation is

`(dy)/(dx)=4x^3\lbrack\frac{x^4}{\sqrt[]{1-x^8}}+\sin ^(-1)(x^4)\rbrack`