Question

dr/dx for r=3secxcscx

Answer 1

\[\Large\bf\sf r=3\sec x \csc x\]Product rule:\[\Large\bf\sf r'=\color{royalblue}{(3\sec x)'}\csc x+3 \sec x \color{royalblue}{(\csc x)'}\]

Answer 2

Need to take the derivative of the blue parts.
Do you remember derivative of secant and cosecant?

Answer 3

If not, you can convert things to sines and cosines :o

Answer 4

you have it backwards

Answer 5

It's addition, there is no backwards with addition.

Answer 6

product rule take derivative of the second term first and i tried changing it like that im just having trouble simplifying

Answer 7

\[\Large\bf\sf r'=\color{orangered}{(3\sec x \tan x)}\csc x+3 \sec x \color{orangered}{(-\csc x \cot x)}\]Is this what you got for your derivatives?

Answer 8

3secx*-cscxcotx+cscx*3secxtanx

Answer 9

yup

Answer 10

where do i go from there

Answer 11

Ok looks good.
I guess from here you need to convert everything to sines and cosines.
It should simplify down quite a bit.

Answer 12

the answer is 3sec^2 x -3csc^2x so im wondering how to get there without changing to sine cosine

Answer 13

You must change to sines and cosines.
There might be another fancy way to do it... I'm not sure..
But this works out really nicely.

Answer 14

ok well which identities are you using to get there. ?

Answer 15

\[\Large\bf\sf \sec x=\frac{1}{\cos x},\qquad\qquad \csc x=\frac{1}{\sin x}\]
\[\Large\bf\sf \tan x=\frac{\sin x}{\cos x},\qquad\qquad \cot x=\frac{\cos x}{\sin x}\]

Answer 16

yea everything eventually cancels and you can convert back secx and cscx eh

Answer 17

Well not everything,
but yes there will be some nice cancellations :)

Answer 18

\[\frac{ 1 }{ \cos x}=(\cos x)^{-1}\]

Just a nice thing to keep in mind if you hate the directionality of the quotient rule and are comfortable with the common chain rule.

Answer 19

Not to be confused with \(\cos^{-1}x\)