Please, if someone can help me with this, i will be indebted to them. Im just..having way too much trouble with this problem. If someone can show the intermediate steps, it would be wildly appreciated (and what rules theyre using)
is that tan^-1 literally the reciproce of tan or the arctan
Okay mate it's pretty simple
you use the derivative formula for inverse tan and the chain rule
>>7639964
Are you asking why the derivative of arctan(f(x)) is
[math]
\frac{f'(x)}{1+(f(x))^2}
[/math]
or what?
shut off steam you ding dong
well, how in the second line did they get (1+(x/(x^2+1)^(1/2))?
>>7639964
Use the chain rule for y = atanx
y' = 1/(1+x^2)
There you go anon. Feel free to ask questions.
>>7639985
When has anyone ever used the reciprocate of tan instead of cot
>>7639964
Okay OP, the derivative of [math] tan^{-1} [/math](x) is [math] \frac{1}{1+x^2} [/math]. So set u=x+[math] \sqrt{x^2+1} [/math], take the derivative and then use chain rule (multiply by the derivative of u). Recall that [math] \sqrt{x^2+1} [/math] is just [math] (x^2+1)^{1/2} [/math], so bring down the 1/2 and the exponent becomes -1/2. Then we use chain rule again and multiply by the derivative of [math] x^2 +1 [/math] and get 2x, which cancels the 1/2 from before. This leaves us with line two from the pic. To get to line 3, you just expand the exponent on the bottom using FOIL, and turning the 1 inside the parentheses into [math] \frac{ \sqrt{x^2+1}}{ \sqrt{x^2+1}} [/math] and combine the fractions. To get to line 4, pull a factor of 2 out of the bottom of the first fraction then multiply the two fractions. To get to line 6, you multiply every term in the bottom by [math] \sqrt{x^2+1} [/math] which gives us [math] 2[ \sqrt{x^2+1} +x^2 \sqrt{x^2+1} +x \sqrt{x^2+1} ^2] [/math] on the bottom. The [math] \sqrt{x^2+1} ^2 [/math] becomes [math] x^2+1 [/math] and the other terms have a factor of [math] \sqrt{x^2+1} [/math] in common that can be pulled out leaving [math] \sqrt{x^2+1} (1+x^2) [/math]. Now we see that there is a common factor of [math] (x^2+1) [/math] in the bottom, so we pull it out leaving [math] (x^2+1) ( \sqrt{x^2+1} +x) [/math]. The two terms cancel out leaving you with the final answer. Hope this helped, let me know if you have any questions.
>>7640044
>When has anyone ever used the reciprocate of tan instead of cot
Uh, I would desu. I much prefer saying arctan than tan^-1.
Doesn't really matter though, trig is shit.
>>7640050
>Uh, I would desu.
>Doesn't really matter though, trig is shit.
confirmed pleb gtfo. Also, when people read tan^-1 they don't say "tan to the minus one" they call it arctan. If you actually write out arctan every time you use it instead of writing tan^-1, you're just wasting energy and time.
>>7640060
>you're just wasting energy and time.
Yes, I'm so super busy I must make time by using a slightly less ambiguous notation. I need to churn out solutions by the minute, no time to use notation that makes sense in the context.
You guys don't know how much this means to me. Im so stressed and have been struggling so much in college with this. You guys are amazing <3 God bless and have a great day
>>7640065
If you aren't super busy, you're wasting time and wasting your life. Get the fuck off /sci/ and actually contribute something to the world. You'll thank me when all those milliseconds add up.
>>7640097
>person on /sci/ tells me to stop wasting my time with /sci/
>>7640103
>person with lung cancer tells you to stop smoking cigarettes
>lol but ur doing it screw off i do what i want :^)
>>7640118
I'm sure you have many more nuggets of wisdom to drop on me.
>>7640123
And I'm sure you have other threads you want to shit up. I'm gonna get the rest of next week's homework finished up then go bang my girlfriend. I'll be back here in a minute 30 if you want to continue our discussion of your waste of existence.
>>7640131
Wait, didn't you just admit that you waste your time on /sci/ by analogy to being a cancer patient?
Why are you all banging-whores-while-doing-homework now?
>>7640010
Can you learn to write neater.
