Thread: Derivative of log

Find where y'=0 given that y=10^xlogx-10x

Originally Posted by loghater

Find where y'=0 given that y=10^xlogx-10x

The derivative is .
The solution by inspection is clear.

How do you get x? I got to the point where (10^xlogx-10x)/logxln10x=0

Originally Posted by loghater

How do you get x? I got to the point where (10^xlogx-10x)/logxln10x=0

That derivative is not correct.

Originally Posted by loghater

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.

Originally Posted by loghater

Find where y'=0 given that y=10^xlogx-10x

As it is written, there is no way to tell whether the original function is or 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.