Question

Show that the derivative of 2*sqrt[tan(x)+sec(x)] = sec(x)*sqrt[tan(x)+sec(x)]
My teacher said to use power rule, then trig and then write the derivative in a fractional radical form & rationalize. I don't know if that helps, but yeah.
So far I have:
[tan(x)+sec(x)]^-1/2[sec(x)^2+tan(x)sec(x)] but i don't know where to go from there to get the answer

Answer 1

hmm i think your teacher meant to say chain rule :)
no 2 functions are being multiplied in this case

Answer 2

oh yeah she meant chain rule, but she usually combines chain and power to the same rule, but yeah she meant chain!

Answer 3

ok so far you are correct....now factor out a sec(x) and simplify like terms by combining exponents

Answer 4

\[\rightarrow \sec(x)[\sec(x)+\tan(x)]*[\sec(x)+\tan(x)]^{-1/2}\]

Answer 5

Hmm okay, so I get how you factor out the sec(x). but im a little lost on which terms I should combine because im guessing i shouldnt combine sec(x) with sec(x)+tan(x) because that would be redundant. would i combine [sec(x)+Tan(x)] with [sec(x)+tan(x)]^-1/2?

Answer 6

get rid of that rational exponent
the derivative of \(\sqrt x\) is \(\frac{1}{2\sqrt x}\)

Answer 7

so by the chain rule, the derivative of
\[2\sqrt{\tan(x)+\sec(x)}\] is
\[\frac{\sec^2(x)+\sec(x)\tan(x)}{\sqrt{\tan(x)+\sec(x)}}\]

Answer 8

factor out the \(\sec(x)\) and get
\[\sec(x)\frac{\sec(x)+\tan(x)}{\sqrt{\sec(x)+\tan(x)}}\]

Answer 9

then it is a fact that \(\frac{a}{\sqrt{a}}=\sqrt{a}\)

Answer 10

Oooooooohhhh thank you soooooo much! I kept getting fairly close, but if I knew that \[\frac{ a }{ \sqrt{a} }\] = sqrt a, then it would have made things much easier! Thanks!!