1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Formatting tips meta.math.stackexchange.com/q/5020/346279 $\endgroup$ – Aakash Kumar Aug 5 '16 at 15:05
  • $\begingroup$ 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 $$\,y' = 2\,\frac{x^{\log_{10} x}\log_{10} x}x=2\,x^{\log_{10}(x)-1}\log_{10} x$$ $\endgroup$ – Raymond Manzoni Aug 5 '16 at 15:39
0
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.