Find where y'=0 given that y=10^xlogx-10x
Important question- is "log x" the common logarithm (base 10) or the natural logarithm (base e)? Normally, for a calculus problem I would assume natural logarithm, but the presence of "10^x" makes me ask.
I presume you know that the derivative of -10x is -10.
10^x log x is a product so use the product rule: (10^x log x)'= (10^x)' log x+ 10^x (log x)'.
The derivative of 10^x is (10^x)'= (10^x) ln x and the derivative of log x is
1) 1/(x ln(10)) if it is common logarithm.
2) 1/x, if it is natural logarithm.
As it is written, there is no way to tell whether the original function is $\displaystyle \displaystyle \begin{align*} y = 10^x\log{x} - 10x \end{align*}$ or $\displaystyle \displaystyle \begin{align*} y = 10^{x\log{x}} - 10x\end{align*}$ or even something else. The OP needs to clarify, and in future, needs to use brackets or at the very least, some space, where they are needed, or even better, to learn some basic LaTeX.