Question

Derivative of 7/sin(x) + 1/cot(x)

Answer 1

I don't know what I did wrong but I got the answer as = -7cosx + csc2x

Answer 2

of am I supposed to use chain rule?

Answer 3

\[\frac{ -7 }{ \cos^2 x } +\sec^2x\]
The derivative as the first part is sinx to the negative one power in the numerator times seven. when you bring the negative one down the seven becomes negative and then you have sine x to the negative second power. The derivative of the inside becomes cosx to the negative second which is cos sqared in the denominator. then add that to the derivative of tan because tan is 1 over cot. The derivative of tanx is sec squared x

Answer 4

YOu can probably use other trig identities to simplify this even further but I would have to try it

Answer 5

I made a mistake this is actually\[\frac{ -7 }{ \sin^2 x } * cosx +\sec^2x\]
because yes, you should use the chain rule. Sorry I messed upon the first answer

Answer 6

so does that make sense @thuyvy ?

Answer 7

yes I got it

Answer 8

k :)

Answer 9

hello, are you still there?

Answer 10

you can type your formula into wolframalpha

Answer 11

thanks

Answer 12

They dont show me step by step, can you help

Answer 13

how did you get tan for the derivative ?

Answer 14

1/cotx is tanx

Answer 15

and 1/sinx is cscx

Answer 16

this will do it, thanks so much :)