Question

Math calculus find the first derivative of the given function. Q2 only

Answer 1

Q2) We have to find the derivative of the function y = (ln x)^x.

We can start using the property of the derivative of an potential and exponential function:

`(d(u^v))/(dx)=u^v\cdot(d\lbrack\ln (u)\cdot v\rbrack)/(dx)`

In this case, u = ln(x) and v = x, so we can solve it as:

`\begin{gathered} (dy)/(dx)=\ln (x)^x\cdot(d\lbrack\ln (\ln (x))\cdot x\rbrack)/(dx) \\ (dy)/(dx)=\ln (x)^x\cdot\lbrack\ln (\ln (x))\cdot1+x(d(\ln (\ln (x)))/(dx)\rbrack \\ (dy)/(dx)=\ln (x)^x\cdot\lbrack\ln (\ln (x))\cdot1+x\cdot(1)/(\ln(x))\cdot(1)/(x)\rbrack \\ (dy)/(dx)=\ln (x)^x\cdot\lbrack\ln (\ln (x))+(1)/(\ln (x))\rbrack \end{gathered}`

Answer: dy/dx = ln(x)^x * [ln(ln(x)) + 1/ln(x)]