Putting `sqrta=tan alpha and sqrtx=tan theta`, we get
`y=tan^(-1)((tan alpha-tan theta)/(1+tan alpha tan theta))=tan^(-1){tan(alpha-theta)}=(alpha-theta).`
`thereforey=(alpha-theta)rArry=tan^(-1)sqrta-tan^(-1)sqrtx`
`rArr(dy)/(dx)=(d)/(dx) (tan^(-1)sqrta)-(d)/(dx)(tan^(-1)sqrtx)`
`={0-(1)/((1+x)).(1)/(2)x^(-1//2)}=(-1)/(2sqrtx(1+x))`.
Hence, `(dy)/(dx)=(-1)/(2sqrtx(1+x)).`
