# chain rule help!

• Sep 6th 2008, 12:21 AM
jvignacio
chain rule help!
differentiate

$

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 AM
kalagota
Quote:

Originally Posted by jvignacio
differentiate

$

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

well, chain rule says that if $h(x)=f(g(x))$, then $h'(x)=f'(g(x))\cdot g'(x)$ of course in the assumption that f and g are differentiable..

so

your $f'(x) = \left[\sec^2{\left(\frac{x}{73}\right)}\right] \cdot \frac{1}{73}$
• Sep 6th 2008, 12:51 AM
jvignacio
Quote:

Originally Posted by kalagota
well, chain rule says that if $h(x)=f(g(x))$, then $h'(x)=f'(g(x))\cdot g'(x)$ of course in the assumption that f and g are differentiable..

so

your $f'(x) = \left[\sec^2{\left(\frac{x}{73}\right)}\right] \cdot \frac{1}{73}$

so the differentiation of x/73 is 1/73?
• Sep 6th 2008, 01:11 AM
11rdc11
Yes, $\frac{1}{73}$ is just a constant multiplied to the variable x.