Question

dy/dx (1/a)arcsec(u/a)

u=function
a=constant

Answer 1

\[\Large f(x)=\frac{ 1 }{ a }\sec ^{-1}\frac{ u }{ a }\]
Find f'(x)

Answer 2

u is the function and a is a constant btw

Answer 3

Yeah, ln(u)u' but I'm trying to work backwards from a formula

Answer 4

ln(x)

Answer 5

Hold on

Answer 6

\[\frac{ d }{ dx }\frac{ 1 }{ a }\sec ^{-1}\frac{ u }{ a }\]

Answer 7

Hint: 1/a is just a constant coefficient. What could we do with it to simplify the differentiation?

Answer 8

If that's unclear, start by finding \[\frac{ d }{ dx }\sec ^{-1}x~ for~ practice.\]

Answer 9

If you're unsure, see the tables in http://www.math.com/tables/derivatives/tableof.htm

Answer 10

I'm trying to get to this formula by differentiating
\[\Large \int\limits\limits_{}^{}\frac{1}{u}\frac{ du }{\sqrt{u^2-a^2} }=\frac{1}{a}arcsec\frac{\left| u \right|}{a}+C\]

Answer 11

What would happen were you to differentiate the left side of the equation immediately above?

Answer 12

As for the right side: start out with the simpler \[y=arcsec x\]with the goal of finding the derivative of y with respect to x.

Answer 13

If y = arcsec x, then x=sec y.

Taking the derivative of both sides, \[\frac{ dx }{ dx }=\frac{ d }{ dx }\sec y\]

Answer 14

Can you finish this differentiation? Hint: use the chain rule last.

Answer 15

1=y'/abs (y)*(y^2-1)

Answer 16

1/abs (y)*(y^2-1)=y'
And now plug y back in?

Answer 17

that's right...solve for (dy/dx). I'd bet that your (dy/dx) is similar to the left side of your original equation.

Think about what we have done so far. What more remains to do? Note that I used x=cos y as an example; now that you have correctly found (dy/dx), you can apply the same technique to find the derivative of your function with argument (u/a). OK?

Answer 18

Ok but what I'm stuck with is plugging my argument in for y. I get 1/abs (abs (x)/a)*sqrt(absx)/a) the x ' are supposed to be U ' but yeah I get stuck

Answer 19

Just a sec, please. I'm going to write this out on paper for myself before trying to explain further what needs to be done.

Answer 20

I have to go ill come ba c k to this

Answer 21

Let's see if the following makes sense to you:

Given y=arcsec x, x-sec y.

Goal: Find (dy/dx).

\[\frac{ d }{ dx }[x=\sec y]\rightarrow \frac{ dx }{ dx }=\sec y \tan y \frac{ dy }{dx }\]

Answer 22

Algebraic manipulation results in \[1=\sec y \tan y \frac{ dy}{ dx}\]

Answer 23

or\[\frac{ dy }{ dx }=\frac{ 1 }{ \sec y \tan y }\]

Answer 24

Review this work. What would you do next? We're almost "there."

Answer 25

Sub the y back in?

Answer 26

I got it now