Question

Help test tomorrow!
Derivative of y=csc^-1(secx), 0<=x<(pi/2)

Answer 1

i posted this a while ago, but no one can help me

Answer 2

i tried to use the rules and stuff, but i got lost somewhere and i don't know where to go from there

Answer 3

did you try chain rule

Answer 4

does anyone know how to do this?

Answer 5

yeah, i used the chain rule, but i got lost somewhere

Answer 6

i think i simplified wrong

Answer 7

try to work it implicitly

Answer 8

take \(\rm \csc^{-1}\) both sides

Answer 9

I got
\[y \prime = -\frac{ 1 }{ secx \sqrt{\sec^2x-1} }*secx*tanx*secx*tanx\]

Answer 10

ok ill try implicitly

Answer 11

\[\rm y=\csc^{-1}(secx) \implies \csc y = \sec x\]

differentiate both sides

Answer 12

why is it cscy and not csc^-1y if you are taking csc^-1 of both sides?

Answer 13

oh never mind

Answer 14

ok, so i got
\[-\csc(y) \cot(y) \frac{ dy }{ dx } = secxtanx\]

Answer 15

\[\csc ^{-1}(\sec (x))\]

Answer 16

is this the eq.

Answer 17

yeah

Answer 18

now i got
\[\frac{ dy }{ dx } = -\frac{ \sec(x)\tan(x) }{ \csc(y)\cot(y) }\]

Answer 19

but that doesn't look right

Answer 20

i know the answer is -1, but i don't understand how to go from there to -1

Answer 21

did i also say that it is from \[0 \le x<\frac{ \pi }{ 2 } \]

Answer 22

where does that come in?

Answer 23

\[\rm y=\csc^{-1}(secx) \implies \csc y = \sec x\]

\[\rm \implies \sin y = \cos x\]

yes ?

Answer 24

ok, yeah, im following

Answer 25

isn't differentiating sin and cos simpler than csc and sec ?

Answer 26

\[\rm y=\csc^{-1}(secx) \implies \csc y = \sec x\]

\[\rm \implies \sin y = \cos x\]

differentating both sides implicitly and solving dy/dx gives you

\[\rm \dfrac{dy}{dx} = -\dfrac{\sin x}{\cos y}\]

Answer 27

ok, i got that

Answer 28

next, lets draw a triangle since we know that the angle is in first quadrant

Answer 29

but how do you work with that y

Answer 30

ok

Answer 31

like that?

Answer 32

Perfect!

Answer 33

find the value of \(\rm \cos y\) and plug it in

Answer 34

is that the value of the third side?

Answer 35

technically yes, but it simplifies

Answer 36

\[\rm 1-\cos^2x = \sin^2x\]

Answer 37

ok

Answer 38

\[cosy = \frac{ -sinx }{ \frac{ dy }{ dx } }\]
right?

Answer 39

and then dy/dx is just -sinx/cosy

Answer 40

\[cosy = \frac{ -sinx }{ \frac{ -sinx }{ cosy } }\]

Answer 41

ok, something went wrong

Answer 42

ok, you know what, thanks for the help, but i have to go to bed. at least i know to study implicit differentiation before my test. thanks for your time

Answer 43

the answer for your q is (-1)..... take the D of arccsc the D of sec

Answer 44

the -tan(x)/sqr(sec^2-1)=-tan(x)/tan(x)

Answer 45

\[\frac{ -1 }{ \left| \sec(x) \right|\sqrt{\sec ^{2}}(x)-1 }*\sec (x)\tan (x)\]