# Derivative of logs in exponent

I don't understand this question. I realize that

$$\log_A x = \ln X/\ln A,$$

but when I substitute that and take the derivative, I get

$$y' = e^{\left(\ln x\right)^2/10}\cdot\frac{2\ln x}{\ln 10}\cdot\frac{1}{x}.$$

How do I continue from here? What is the derivative of $$f(x) = x^{\log_{10} x}?$$

EDIT: I see how Raymond got his answer but my book says this is $$y' = x^{\log_{10} x}\cdot\frac{ln 10\cdot\log_{10} x + ln x} {x\cdot\ln 10}.$$ So how can one get this answer?

• Formatting tips meta.math.stackexchange.com/q/5020/346279 – Aakash Kumar Aug 5 '16 at 15:05
• A '$\ln$' is missing before the $10$ in your expression for $y'$ : $$\,y' = e^{(\ln x)^2/\ln 10}\left(2\frac{\ln x}{\ln 10}\right)\frac 1x$$ else your expression is correct and you may revert the '$\log_A$'operation and write this as $$\,y' = 2\,\frac{x^{\log_{10} x}\log_{10} x}x=2\,x^{\log_{10}(x)-1}\log_{10} x$$ – Raymond Manzoni Aug 5 '16 at 15:39

## 1 Answer

A '$\ln$' is missing before the $10$ in your expression for $y'$ : $$\,y' = e^{\large{(\ln x)^2/\ln 10}}\left(2\frac{\ln x}{\ln 10}\right)\frac 1x$$ else your expression is correct and you may revert the '$\log_A$'operation and write this as (using $\ln(x)^2/\ln 10=\log_{10}(x)\,\ln(x)$) : $$\,y' = 2\,x^{\large{\,\log_{10} x}}\frac{\log_{\,10} x}x=2\,x^{\large{\,\log_{10} x}-1}\log_{\,10} x$$

ADDITION

Concerning your book's answer it is right but a little 'poor' since : \begin{align} y' &= x^{\large{\,\log_{10} x}}\cdot\frac{\ln 10\cdot\log_{10} x + \ln x} {x\cdot\ln 10}\\ &= x^{\large{\,\log_{10} x}}\cdot\frac{\log_{10} x + \frac{\ln x}{\ln 10}} {x}\\ &= x^{\large{\,\log_{10} x}}\cdot\frac{2\,\log_{10} x} {x}\\ \end{align} as previously...