यदि `y=btan^(-1)(x/a+tan^(-1)y/x)` हो,तो `(dy)/(dx)` ज्ञात कीजिए ।

Explanation

`y=btan^(-1)(x/a+tan^(-1)y/x)`
`rArry/b=tan^(-1)(x/a+tan^(-1)y/x)`
`rArrtany/b=x/a+tan^(-1)y/x`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`d/(dx){tan(y/b)}=d/(dx)(x/a)+d/(dx){tan^(-1)(y/x)}`
`rArrsec^(2)(y/b).1/b(dy)/(dx)=1/a+1/(1+(y/x)^(2))`,
`((x(dy)/(dx)-y.1)/(x^(2)))`
`rArr1/bsec^(2)(y/b)(dy)/(dx)=1/a+(x^(2))/(x^(2)+y^(2)).((x(dy)/(dx)-y))/(x^(2))`
`rArr1/bsec^(2)(y/b)(dy)/(dx)=1/a+x/(x^(2)+y^(2))(dy)/(dx)-y/(x^(2)+y^(2))`
`rArr[1/bsec^(2)(y/b)-x/(x^(2)+y^(2))](dy)/(dx)=1/a-y/(x^(2)+y^(2))`
`rArr(dy)/(dx)=(1/a-y/(x^(2)+y^(2)))/(1/bsec^(2)(y/b)-x/(x^(2)+y^(2)))`