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Differentiate the function `(sin x)^x+ sin^1 x` with respect to x. |
Answer» `"Let " y = ("sin"x)^(x) + "sin"^(-1) sqrt(x)` `"Let " u = ("sin"x)^(x) "and " v = "sin"^(-1) sqrt(x)` `therefore y = u+v` `rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " " .....(1)` `"Now", u = ("sin" x)^(x)` `rArr "log"u = "log"("sin" x)^(x) = x "log " "sin"x` `rArr (1)/(u) (du)/(dx) = x (d)/(dx) "log sin" x + "log sin"x (d)/(dx)x` `rArr (du)/(dx) = u[x* ("cos"x)/("sin"x) + "log sin" x * 1]` `rArr (du)/(dx) = ("sin"x)^(x) [x "cot" x + "log sin"x]` `"and " v="sin"^(-1) sqrt(x)` `rArr (dv)/(dx) = (d)/(dx)"sin"^(-1) sqrt(x)` `= (1)/(sqrt(1-(sqrt(x))^(2))) (d)/(dx)sqrt(x) = (1)/(2sqrt(x)(1-x))` From equation (1) `(dy)/(dx) = ("sin" x)^(x)[x "cot"x + "log sin"x] + (1)/(2sqrt(x)(1-x))` |
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