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Find derivative of `cos(logx+e^x)`, `x >0`w.r.t. to `x` |
Answer» `"Let" y = "cos" ("log" x +e^(x))` `rArr (dy)/(dx) =(d)/(dx)"cos" ("log" x + e^(x))` `=-"sin"("log" x + e^(x))(d)/(dx)("log" x +e^(x))` `=-"sin"("log" x +e^(x))((1)/(x)+e^(x))` `=(-"sin" ("log" x + e^(x))(1 + x * e^(x)))/(x)` |
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