1. ## Derivatives

find the dy/dx of y=4(sec x+ tan x)^1/2, idk how to put root...

2. $4(sec x+ tan x)^{1/2} = 4 \sqrt{sec(x)+tan(x)}$

Use the Chain Rule.

3. I know i need to use Chain Rule but i don't know how!!!

4. I'm no expert at this (posted a chain rule question yesterday actually) but I think you want to take dy/du of 4u^1/2 and du/dx of u=(sec x + tan x) and multiply. Then substitute (sec x + tan x) back into the equation for u.

I'll stick around for the answer to this question because I'm not confident in my work. When I try to solve I get an ugly answer, which may be true...but I'm not sure.

5. Chain rule:

$(f \cdot g)'(x) = f'[g(x)] \cdot g'(x)$

Leave the function as a power $y= 4 (\sec x+ \tan x)^{1/2}$ and let $\sec x+ \tan x = g(x)$

Can you show us your attempt at it?

6. $\displaystyle \left(\sqrt{f(x)}\right)' = \frac{f'(x)}{2\sqrt{f(x)}}$

7. i don't understand what is the derivative of y=4(sec x+ tan x)^1/2.............................
is it 7/2(secx+tanx)^-1/2??? pretty much this is why iam asking this question..

8. Originally Posted by kevinoh92
i don't understand what is the derivative of y=4(sec x+ tan x)^1/2.............................
is it 7/2(secx+tanx)^-1/2??? pretty much this is why iam asking this question..
differentiate 4Sqrt&#91;&#40;1&#47;Cos&#91;x&#93;&#43; Tan&#91;x&#93;&#93; - Wolfram|Alpha

Click on Show steps.

9. Thanks ,mr fantastic