`d/(dx)[tan^(-1)((sqrt(x)(3-x))/(1-3x))]=` `1/(2(1+x)sqrt(x))` (b) `3/((1+x)sqrt(x))` `2/((1+x)sqrt(x))` (d) `3/(2(1+x)sqrt(x))`
`d/(dx)[tan^(-1)((sqrt(x)(3-x))/(1-3x))]=`
`1/(2(1+x)sqrt(x))`
(b) `3/((1+x)sqrt(x))`
`2/((1+x)sqrt(x))`
(d) `3/(2(1+x)sqrt(x))`
A. `(1)/(2(1+x)sqrt(x))`
B. `(3)/((1+x)sqrt(x))`
C. `(2)/((1+x)sqrt(x))`
D. `(3)/(2(1+x)sqrt(x))`
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1 Answers
`(d)/(dx)(tan^(-1)""(sqrt(x)(3-x))/(1-3x))`
`(d)/(dx)(tan^(-1)""((tan theta(3-tan^(2)theta)))/(1-3tan^(2)theta))`
`("putting "sqrt(x)=tan theta or theta=tan^(-1)sqrt(x))`
`=(d)/(dx)(tan^(-1)""((3tantheta-tan^(3)theta))/(1-3tan^(2)theta))`
`=(d)/(dx)[tan^(-1)(tan 3theta)]=(d)/(dx)(3theta)`
`=(d)/(dx)[3tan^(-1)sqrt(x)]=(3)/(2sqrt(x)(1+x))`
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Answered