Find the Derviv (inverse tan prob)

[calcstruggles2013]

find h'(x) for h(x) = 15 arctan

(

x ​ √ ​ 3

)

sorry that this looks so weird but it is 15arctan(x/(3^1/2))

I know that y’ of arctan is 1/1+x^2 but im having trouble know that the argument is x divided by the sqroot of 3. I’m not quite sure how one would incorporate that into the simpler dervivative.
I know that when a constant is in front of the function you would just use the products rule,
Thus I would need to solve for 15* (arctan(x/sqroot(3)))’

I know that arctan(a/x)’ is arctan(-a/x^2 + a^2) so using this I got
15*((-sqrt(3))/(sqrt(3))^2+x^2)
Lon capa says im wrong. I’m not sure if there should be a negative in front of the square root or not either.

 

[srmichael]

find h'(x) for h(x) = 15 arctan

(

x ​ √ ​ 3

)

sorry that this looks so weird but it is 15arctan(x/(3^1/2))

I know that y’ of arctan is 1/1+x^2 but im having trouble know that the argument is x divided by the sqroot of 3. I’m not quite sure how one would incorporate that into the simpler dervivative.
I know that when a constant is in front of the function you would just use the products rule,
Thus I would need to solve for 15* (arctan(x/sqroot(3)))’

I know that arctan(a/x)’ is arctan(-a/x^2 + a^2) so using this I got
15*((-sqrt(3))/(sqrt(3))^2+x^2)
Lon capa says im wrong. I’m not sure if there should be a negative in front of the square root or not either.

Let \(\displaystyle u=\frac{x}{\sqrt{3}}\)

And you don't have to do product rule if you have a constant time a function. The product rule is for when you have two functions multipled by each other. So just take the derivative of arctan(u) like normal then multiply the answer by 15.

 

[calcstruggles2013]

woohoo finally got it thanks
15*1.73205080756888/(3+x^2)