0
$\begingroup$

Can you help me to find a second derivative of func?

$f(x) = \lg(x) + 5-x$

If you can, please , explained if for me.

Thank you.

$\endgroup$
  • $\begingroup$ $\log$, as in base $10$ or base $e$? $\endgroup$ – S.C.B. Feb 19 '17 at 16:54
  • $\begingroup$ I hope you know that the derivative of a constant is zero and that of $\log x $ is $1/x $. $\endgroup$ – Rohan Feb 19 '17 at 16:54
  • $\begingroup$ yes, lg @S.C.B. $\endgroup$ – Vadim Marchenko Feb 19 '17 at 16:55
  • $\begingroup$ lg(x) + 5 -x ; @Rohan $\endgroup$ – Vadim Marchenko Feb 19 '17 at 16:55
0
$\begingroup$

We have $$f (x) = \log_{10} x + 5-x $$ $$\Rightarrow f'(x) = \frac {\mathrm {d}}{\mathrm {d}x}(\frac {\log_e x}{\log_e 10} + 5-x) = \frac {1}{x\log_e 10} + 0 -1 =\frac {1}{x\log_e 10}-1$$ $$\Rightarrow f''(x) = \frac {\mathrm {d}}{\mathrm {d}x}(\frac {1}{x\log_e 10}-1) = \frac {-1}{x^2\log_e 10} + 0 = -\frac {1}{x^2\log_e 10}$$

We have just used the idea expressed in my commemt above. Also $\log_e 10 = \ln 10$, can be taken out of the expression while taking the derivative and does not affect it. Hope it helps.

$\endgroup$
0
$\begingroup$

$$f(x)=\log(x)+5-x$$

so $$f'(x)=\frac{1}{xlog(10)}-1$$

so $$f''(x)=-\frac{1}{log(10)x^{2}}$$ using the quotient rule and that the derivative of a constant is zero

$\endgroup$
  • $\begingroup$ f(x)=lg(x)+5−x ; lg, not ln $\endgroup$ – Vadim Marchenko Feb 19 '17 at 16:59
  • $\begingroup$ @VadimMarchenko Log base 10? $\endgroup$ – Quality Feb 19 '17 at 17:00
  • $\begingroup$ yes, log base 10 $\endgroup$ – Vadim Marchenko Feb 19 '17 at 17:02

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.