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`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))` |
Answer» `(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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