निम्नलिखित फलनों का अवकलज ज्ञात कीजिए-
`tan^(-1)[(x^(1//3)+a^(1//3))/(1-x^(1//3)a^(1//3))]`

Explanation

माना `y=tan^(-1)[(x^(1//3)+a^(1//3))/(1-x^(1//3)a^(1//3))]`
अब `x^(1//3)=tanA,a^(1//3)=tanB` रखने पर,
`thereforeA=tan^(-1)x^(1//3)` "" `B=tan^(-1)a^(1//3)`
`thereforey=tan^(-1)[(tanA+tanB)/(1-tanAtanB)]`
`rArry=tan^(-1)[tan(A+B)]=A+B`
`rArry=tan^(-1)x^(1//3)+tan^(-1)a^(1//3)`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`(dy)/(dx)=d/(dx)[tan^(-1)x^(1//3)]+d/(dx)[tan^(-1)a^(1//3)]`
`=1/(1+(x^(1//3))^(2)).d/(dx)(x^(1//3))+0`
`=1/(1+x^(2//3)).1/3x^(-2//3)`
`=1/(3x^(2//3)(1+x^(2//3)))`.