chain rule help!

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September 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
September 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 $f'(x) = \left[\sec^2{\left(\frac{x}{73}\right)}\right] \cdot \frac{1}{73}$
September 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?
September 6th 2008, 01:11 AM
11rdc11

Yes, $\frac{1}{73}$ is just a constant multiplied to the variable x.

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