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If `y=(x^x)^x` then `(dy)/(dx)` isA. `y[x^(x)(log ex)log x+x^(x)]`B. `y[x^(x)(log ex)log x+x]`C. `y[x^(x)(log ex)log x+x^(-1)]`D. `y[x^(x)(log_(e)x)log x+x^(-1)]` |
Answer» `y=""_(x)(x^(x))` `"or "log y= x^(x) log x` `"or "(1)/(y)(dy)/(dx)=(dz)/(dx)log x +(1)/(x)z("where "x^(x)=z)` `"or "(dy)/(dx)=x^((x^(x)))[x^(x)(log ex)log x + x^(x-1)]" "(because(dz)/(dx)=x^(x) log ex)` |
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