Question

What is the derivative of y = ln ( sec ( √ X ) + tan ( √ X ) ) ?

Answer 1

Answer:

dy / dx= sec √x / (√2x)

Step-by-step explanation:

y=ln ( sec (√ x ) + tan ( √ x ) ) ?

let √x=t

y= ln ( sec t +tant)

also let  z = sect + tan t

y = ln z

dy/dz= 1/z.................(1)

dz/dt =sect tan t +sec² t .....(2)

t = √x

dt / dx =1/2 x(1/2-1)=1/2*1/√x = 1 /√2x......(3)

Now we know that

dy/dx = (dy/dz * dz/dt *dt/dx)

Putting all the values we get

dy/dx = (1/z)( sect tan t +sec² t ) ( 1 /√2x)

=( 1/(sect + tan t) ) ( sect tan t + sec²t) / (√2x)

= { sect (tant sect) / { (√2x)(sect + tan t) }

=sec t /(√2x)

dy / dx= sec √x / (√2x)