Always derivatives...
Find the derivative of :
g (x) = arctan {1n [sec √X]}
I'm confuse when I try to resolve the equation
I get derivative step by step, beginning by :
Tanª = 1/ (sec²(v))
{Here I use, v = (sec√x) and a = -1}
Then,
Ln'(v) = 1/v · v'
Next, internal derivative of v :
(sec√x · Tan√x) · ½ · xⁿ
{I use n = - ½ }
So, my entire equation would be :
g(x)' = 1/ (sec²(v) · v'/v · (sec √x · Tan√x)/2 · xⁿ)
{I use a = -½}
Then,
g'(x)' = √x / (sec²(Ln [sec√x]) · sec√x · Tan√x)
But I doubt that is correct and I don't know if is possible more simplification
(I'm sorry for the traslation the terms)
-the derivative of arctan(x) is $\displaystyle \frac{1}{1+x^2}$ ,
in this case: $\displaystyle \frac{1}{log^{2}(\sec(\sqrt{x})+1)}$...(1)
-the derivative of ln(x) is $\displaystyle \frac{1}{x}$ ,
in this case: $\displaystyle \frac{1}{\sec(\sqrt(x))}$....(2)
-the derivative of sec(x) is \sec(x).\tan(x),
in this case $\displaystyle \sec(\sqrt{x}).\tan(\sqrt{x})$...(3)
-finally the sqrt of $\displaystyle \sqrt{x} = \frac{1}{2 \sqrt(x)}$...(4)
multiply 1,2,3,and 4 and simplify
Lets take a look at the derivatives of each of the parts on their own:
$\displaystyle \frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^{2}}$
$\displaystyle \frac{d(ln(x))}{dx}=\frac{1}{x}$
$\displaystyle \frac{d(\sec(x))}{dx}=\sec (x) \tan (x)$
$\displaystyle \frac{d(\sqrt{x})}{dx}=\frac{1}{2\sqrt{x}}$
See if you can use the above to make a composition of functions of derivatives - or just use the derivatives to navigate to an answer.