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

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

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

>>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 GMT No. 16070981 Report

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

Recall:
$\sec (x)=\frac{1}{\cos (x)}$
So
$$\begin{gathered} \cos (x)\tan (x)+\sin (x)\frac{1}{{\cos }^2(x)} \\ \cos (x)\frac{\sin (x)}{\cos (x)}+\sin (x)\frac{1}{{\cos }^2(x)} \\ \sin (x)+\frac{1}{\cos (x)}\frac{\sin (x)}{\cos (x)} \\ \sin (x)+\sec (x)\tan (x) \end{gathered}$$

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

>>16070957
Trivial

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

>>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 GMT No. 16070997 Report

>>16070987
varied

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

>>16070993
>thank you
Anytime.

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

>homework thread

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

>>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 GMT No. 16071035 Report

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