1. ## Find a Derivative

Always derivatives...

Find the derivative of :

g (x) = arctan {1n [sec √X]}

I'm confuse when I try to resolve the equation

2. There's a whole lot of chain rule going on. Let's see what you get.

3. I get derivative step by step, beginning by :

Tan
ª = 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 · x
{I use n = - ½ }

So, my entire equation would be :

g(x)' = 1/ (sec²(v) · v'/v · (sec x · Tanx)/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)

4. This:

g (x) = arctan {1n [sec √X]}
What is that? Is it $tan^{-1}(ln[sec\sqrt{x}])$ or $tan^{-1}\left(\frac{1}{sec\sqrt{x}}\right)$

5. It's true, I'm mistaken...
It's the first equation that you writed (with a logarithm)

6. Originally Posted by AlanaSan
It's true, I'm mistaken...
It's the first equation that you writed (with a logarithm)

-the derivative of arctan(x) is $\frac{1}{1+x^2}$ ,

in this case: $\frac{1}{log^{2}(\sec(\sqrt{x})+1)}$...(1)

-the derivative of ln(x) is $\frac{1}{x}$ ,

in this case: $\frac{1}{\sec(\sqrt(x))}$....(2)

-the derivative of sec(x) is \sec(x).\tan(x),

in this case $\sec(\sqrt{x}).\tan(\sqrt{x})$...(3)

-finally the sqrt of $\sqrt{x} = \frac{1}{2 \sqrt(x)}$...(4)

multiply 1,2,3,and 4 and simplify

7. Originally Posted by AlanaSan
It's true, I'm mistaken...
It's the first equation that you writed (with a logarithm)

Lets take a look at the derivatives of each of the parts on their own:

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^{2}}$

$\frac{d(ln(x))}{dx}=\frac{1}{x}$

$\frac{d(\sec(x))}{dx}=\sec (x) \tan (x)$

$\frac{d(\sqrt{x})}{dx}=\frac{1}{2\sqrt{x}}$

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.