Second derivative of $f(x) = \lg(x) + 5-x$

Question

Can you help me to find the second derivative of

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

If you can, please, explain it to me.

Thank you.

Answer 1

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.

Answer 2

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

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