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Differentiate `x^(xcosx)+(x^2+1)/(x^2-1)`with respect to `x`: |
Answer» `"Let "y = x^(x"cos"x) + (x^(2) + 1)/(x^(2)-1)` `"Let "u = x^(x"cos"x) " and "v=(x^(2)+1)/(x^(2)-1)` `therefore y=u+v` `rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " ".....(1)` `"Now", u=x^(x"cos"x)` `rArr "log" u = "log"(x^(x"cos"x)) = x"cos"x * "log"x` `rArr (1)/(u)(du)/(dx) = x "cos"x * (d)/(dx)"log"x + x"log"x * (d)/(dx)"cos"x + "cos"x * "log"x * (d)/(dx)x` `rArr (du)/(dx) = u(x"cos"x * (1)/(x)-x"log"x * "sin"x + "cos"x * "log"x)` `= x^(x"cos"x) ("cos"x-x"log"x"sin"x + "cos"x"log"x)` `"and "v = (x^(2) + 1)/(x^(2)-1)` `rArr (dv)/(dx) = ((x^(2)-1)(d)/(dx)(x^(2)+1) - (x^(2)+1)(d)/(dx)(x^(2)-1))/(x^(2)-1)^(2)` `= ((x^(2)-1) * 2x - (x^(2)+1) * 2x)/(x^(2)-1)^(2)` `=-(4x)/(x^(2)-1)^(2)` From equation (1) `(dy)/(dx) = x^(x"cos"x)("cos" x - x"log"x"sin"x + "cos"x"log"x)-(4x)/(x^(2)-1)^(2)` |
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