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

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.

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

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.

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

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

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