# Help with Derivatives

by Tom McCurdy
Tags: derivatives
 P: 1,109 My friend asked for some help with derviatives, I said I would explain here then link him Here is how you do it $$\frac {d}{dx} log_b(x) = \frac {1}{xlnb}$$
 P: 1,109 so you have at first $$\frac {d}{dx} log_{10}(10/x) = \frac {1}{\frac{10}{x}ln10}$$ which simplifies to $$\frac {x}{10ln(10}$$ now you may think your done, but you need to remember the chain rule so you have $$\frac {x}{10ln(10} + \frac {d}{dx} \frac {10}{x}$$ so lets take the quotient rule and solve for $$\frac {10}{x}$$ f(x) = 10 f'(x)=0 g(x) = x g'(x)=1 g(x)f'(x)-f(x)g'(x) ---------------- g(x)^2 x*0-10*1 --------- x^2 = $$\frac {-10}{x^2}$$ so then muliply $$\frac {x}{10ln(10)} * \frac {-10}{x^2}$$ and you will get $$\frac {d}{dx} log_{10}(10/x)= \frac {-1}{ln(10)*x}$$