Anonymous at Tue, 12 Mar 2024 23:40:28 UTC No. 16070963

>>16070957
Hint: tanx = (sinx/cosx)

Anonymous at Tue, 12 Mar 2024 23:44:15 UTC No. 16070968

>>16070957
https://en.wikipedia.org/wiki/Trigonometric_functions
Look at the definitions based on hypotenuse, opposite and adjacent.

Anonymous at Tue, 12 Mar 2024 23:51:39 UTC No. 16070981

>>16070957
>Apply product rule
>Then differentiate
>Apply one of the trig identities.

Recall:
[eqn] \sec(x) = \frac{1}{\cos(x)} [/eqn]
So
[eqn]\cos(x) \tan(x) + \sin(x) \frac{1}{\cos^2(x)}\\ \cos(x) \frac{\sin(x)}{\cos(x)} + \sin(x) \frac{1}{\cos^2(x)}\\ \\
\sin(x) + \frac{1}{\cos(x)}\frac{\sin(x)}{\cos(x)}\\ \sin(x) + \sec(x)\tan(x) [/eqn]

Anonymous at Tue, 12 Mar 2024 23:55:34 UTC No. 16070987

>>16070957
Trivial

Anonymous at Tue, 12 Mar 2024 23:57:31 UTC No. 16070993

>>16070963
got that part now, thanks
>>16070981
>>Apply one of the trig identities

>Recall:
sec(x)=1cos(x)

oh my God, shit. I need to brush up on my trig identities. thank you

Anonymous at Tue, 12 Mar 2024 23:58:34 UTC No. 16070997

>>16070987
varied

Anonymous at Tue, 12 Mar 2024 23:59:50 UTC No. 16071000

>>16070993
>thank you
Anytime.

Anonymous at Wed, 13 Mar 2024 00:11:11 UTC No. 16071021

>homework thread

Anonymous at Wed, 13 Mar 2024 00:17:23 UTC No. 16071031

>>16071021
And they will do it for him, for free. Everytime. You fags know you can go get paid for that right?

Anonymous at Wed, 13 Mar 2024 00:18:18 UTC No. 16071035

>>16071021
>>16071031
Better than 99% of threads on /sci/ right now.