|
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]
|