Question

Differentiation of secx(tan√x)? In few steps.

Answer 1

`(dy)/(dx)=secx[(sec^2sqrtx)/(2sqrtx)+tansqrtxtanx]`

Explanation:

`y=secx(tan√x)`
`:.(dy)/(dx)=secxd/(dx)(tansqrtx)+tansqrtxd/(dx)(secx)`
`=secxsec^2sqrtxd/(dx)(sqrtx)+tansqrtxsecxtanx`
`=secxsec^2sqrtx(1/(2sqrtx))+tansqrtxsecxtanx`
`=1/(2sqrtx)secxsec^2sqrtx+tansqrtxsecxtanx`
`(dy)/(dx)=secx[(sec^2sqrtx)/(2sqrtx)+tansqrtxtanx]`

Answer 2

`dy/dx=secxsec^2sqrtxsqrtx/(2x)+tansqrtxsecxtanx`

Explanation:

.

`y=secxtansqrtx`

let `sqrtx=u=x^(1/2), :. x=u^2`

`(du)/dx=1/2x^(-1/2)=1/(2sqrtx)=sqrtx/(2x)`

`d(tansqrtx)=d(tanu)=sec^2udu`

Let's divide both sides by`dx`:

`(d(tansqrtx))/dx=sec^2u((du)/dx)=sec^2sqrtxsqrtx/(2x)`

`dy/dx=secx((d(tansqrtx))/dx)+tansqrtxsecxtanx`

`dy/dx=secxsec^2sqrtxsqrtx/(2x)+tansqrtxsecxtanx`