Y=f(x^2)\\f'(x)=√(5x-1)\\ What is (dy)/(dx)?

Question

`y=f(x^2)\\f'(x)=√(5x-1)\\`

What is
`(dy)/(dx)`?

Answer 1

use chain rule

remember that
`(d)/(dx) g(h(x))=g'(h(x))h'(x)` (mixing Lagrange's notation and Leibniz's notation)

`(dy)/(dx)=`

`(d)/(dx)y=`

`(d)/(dx)f(x^2)=`

if we treat f(x) as g(x) and
`x^2` as h(x) then setup parts

`f'(x)=√(5x-1)` (given) so
`f'(x^2)=√(5x^2-1)`

and h'(x)=
`(d)/(dx)x^2=2x`

`(d)/(dx)y=`

`(√(5x^2-1))(2x)=`

`2x√(5x^2-1)`

answer:
`(dy)/(dx)=2x√(5x^2-1)`

Answer 2

Answer:
`(dy)/(dx)=√(20x^3-4x)`

Step by step:

For better understanding I am using a new variable z instead of original x and then substitute x^2 for z. This makes the use of the chain rule clearer. Let me know if you have questions.

`y=f(z)\\f'(z)=(dy)/(dz)=√(5z-1)\\(dy)/(dx)=(dy)/(dz)(dz)/(dx)\\z=x^2\\y=f(x^2)\\(dy)/(dx)=f'(x^2)(d(x^2))/(dx)\\(dy)/(dx)=√(5x^2-1)\cdot2x\\(dy)/(dx)=√(20x^3-4x)`