Question 709171
{{{ r= root(4, sec(3theta) ) }}}
First of all, when there are radicals in a function (like this one) rewrite them using a fractional exponent instead of the radical before trying to find the function's derivative. (We do this because I've never heard of derivatives of radicals being taught. But we certainly learn how to find derivatives of a power.) So the first thing we do is rewrite the radical:
{{{ r= (sec(3theta))^(1/4) }}}<br>
This is a chain rule problem. At the outermost level we have something raised to the 1/4th power. Inside that we have the sec of something. Inside that we have 3 times theta. (Note: I'm using the very vague term "something" on purpose (as I hope you will understand why as I go through the solution).<br>
We start on the outside and work our way in. The outer most function is something to the 1/4th power. The derivative of this, according to the chain rule, is:
(1/4)*(something #1)^((1/4)-1)*(derivative of something #1)<br>
"Something #1" is: sec(something #2). The "derivative of something #1", according to the chain rule, will then be:
sec(something #2)*tan(something #2)*(derivative of something #2)<br>
"Something #2" is {{{3theta}}}. The "derivative of something #2" is simply 3.<br>
Putting all this together we get:
r' = {{{(1/4)*(sec(3theta))^((1/4)-1)*(sec(3theta)*tan(3theta)*3)}}}
Simplifying...
The 3 at the end and the 1/4 in front make 3/4 and the 1/4 - 1 make -3/4:
r' = {{{(3/4)*(sec(3theta))^(-3/4)*sec(3theta)*tan(3theta)}}}
Since the sec's have the same argument, {{{3theta}}}, we can multiply them. The rule is to add exponents. The (unseen) exponent on the second sec is 1. Adding -3/4 and 1:
r' = {{{(3/4)*(sec(3theta))^(1/4)*tan(3theta)}}}
This may be acceptable as an answer. Or perhaps reverting to radical form would be preferred:
r' = {{{(3/4)*root(4, sec(3theta))*tan(3theta)}}}
or, to ensure there is no confusion about whether the tan is inside or outside the radical:
r' = {{{(3/4)*tan(3theta)*root(4, sec(3theta))}}}