Question

Y=x.arctan(x)^1/2. find dy/dx. pls show steps​

Answer 1

`y=x(\arctan x)^(1/2)`

Use the product rule first:

`(\mathrm dy)/(\mathrm dx)=(\mathrm dx)/(\mathrm dx)(\arctan x)^(1/2)+x(\mathrm d(\arctan x)^(1/2))/(\mathrm dx)`

`(\mathrm dy)/(\mathrm dx)=(\arctan x)^(1/2)+x(\mathrm d(\arctan x)^(1/2))/(\mathrm dx)`

Use the chain rule to compute the derivative of
`(\arctan x)^(1/2)`. Let
`z=(\arctan x)^(1/2)` and take
`u=\arctan x`, so that by the chain rule

`(\mathrm dz)/(\mathrm dx)=(\mathrm dz)/(\mathrm du)(\mathrm du)/(\mathrm dx)`

`(\mathrm dz)/(\mathrm du)=(\mathrm du^(1/2))/(\mathrm du)=\frac12u^(-1/2)`

`(\mathrm du)/(\mathrm dx)=(\mathrm d\arctan x)/(\mathrm dx)=\frac1{1+x^2}`

`\implies(\mathrm d(\arctan x)^(1/2))/(\mathrm dx)=\frac1{2(\arctan x)^(1/2)(1+x^2)}`

So we have

`(\mathrm dy)/(\mathrm dx)=(\arctan x)^(1/2)+\frac x{2(\arctan x)^(1/2)(1+x^2)}`

You can rewrite this a bit by factoring
`(\arctan x)^(-1/2)`, just to make it look neater:

`(\mathrm dy)/(\mathrm dx)=\frac1{2(\arctan x)^(1/2)}\left(2\arctan x+\frac x{1+x^2}\right)`