Help with Derivatives

Tom McCurdy

My friend asked for some help with derviatives, I said I would explain here then link him

Here is how you do it

[tex] \frac {d}{dx} log_b(x) = \frac {1}{xlnb} [/tex]

Tom McCurdy

so you have at first
[tex] \frac {d}{dx} log_{10}(10/x) = \frac {1}{\frac{10}{x}ln10} [/tex]

which simplifies to

[tex] \frac {x}{10ln(10} [/tex]

now you may think your done, but you need to remember the chain rule so you have

[tex] \frac {x}{10ln(10} + \frac {d}{dx} \frac {10}{x}[/tex]

so lets take the quotient rule and solve for [tex] \frac {10}{x}[/tex]

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

=
[tex] \frac {-10}{x^2} [/tex]

so then muliply

[tex] \frac {x}{10ln(10)} * \frac {-10}{x^2} [/tex]

and you will get

[tex]\frac {d}{dx} log_{10}(10/x)= \frac {-1}{ln(10)*x} [/tex]

Tom McCurdy

Any other help you could probably use this website address
http://people.hofstra.edu/faculty/St...s/unit3_3.html