differentiate

$\displaystyle

f(x) = tan(\frac{x}{73})

$

any ideas? i know i have to use the chain rule but confused on the sec^2 and x/73

Printable View

- Sep 6th 2008, 12:21 AMjvignaciochain rule help!
differentiate

$\displaystyle

f(x) = tan(\frac{x}{73})

$

any ideas? i know i have to use the chain rule but confused on the sec^2 and x/73 - Sep 6th 2008, 12:45 AMkalagota
well, chain rule says that if $\displaystyle h(x)=f(g(x))$, then $\displaystyle h'(x)=f'(g(x))\cdot g'(x)$ of course in the assumption that f and g are differentiable..

so

your $\displaystyle f'(x) = \left[\sec^2{\left(\frac{x}{73}\right)}\right] \cdot \frac{1}{73}$ - Sep 6th 2008, 12:51 AMjvignacio
- Sep 6th 2008, 01:11 AM11rdc11
Yes, $\displaystyle \frac{1}{73}$ is just a constant multiplied to the variable x.