Question

Take derivative of y=sin^-1 (3x) +ln(xsecx)^(1/2)

Answer 1

Assuming that your function's first term is\[\sin ^{-1}3x\] find the derivative.

Have you learned the formula for the derivative of the basic inverse sine function? If so, what is \[\frac{ d }{ dx }\sin ^{-1}x~?\]

Answer 2

hmm I don't remember this rule

Answer 3

Once you have that, extend your work to finding the derivative of \[\sin ^{-1}u\]where u is a separate function of x. Use the chain rule.

Answer 4

You'll definitely need to learn the derivative rules for the inverse trig functions. Have you a textbook? Or have you online learning materials that list important derivative formulas, such as this one? Or have you done an internet search for "derivatives of inverse trig functions?"

Answer 5

is it where f'(x)*(1/g'(f(x)) ?

Answer 6

ok, I will read up on this and resubmit my question later

Answer 7

Please refer to:

http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx

You don't have to wade through the whole document. About half way through the doc you'll find a table of the derivatives of the six inverse trig functions.

Answer 8

1 / sqrt(1 - x^2)

Answer 9

sin^-1 and arcsin are equivalent statements ?

Answer 10

Good luck!

Please consider making up your own "cheat sheet" for study purposes. List, for example, all derivative formulas as you learn them. A page or two of summary per section would be a reasonable amount of info to start with.

Yes, the inverse sine can be expressed as arcsin x as well as\[\sin ^{-1}x\]

Answer 11

ok, so is it equal to what I said? :)

Answer 12

Yes,\[\frac{ d }{ dx }\sin ^{-1}x=\frac{ 1 }{ \sqrt{1-x^2} }\]

Answer 13

so sin^-1(3x) -> 1 / sqrt(1 - 3x^2) ?

Answer 14

or do I factor out the 3?

Answer 15

You're on the right track, but parentheses are missing from your result, and you need to apply the Chain Rule:

\[\frac{ d }{ dx }\sin ^{-1}3x=\frac{ 1 }{ \sqrt{1-(3x)^2} }\frac{ d }{ dx }(3x)\]

Answer 16

ok

Answer 17

Please simplify this as far as you can.

Word to the wise: It'd be worth the time and effort required to learn how to use Equation Editor. The reults are so much clearer if you do use that Editor.

Answer 18

Ok, I will work on learning that system. How do I work with the derivative on the second part?

Answer 19

Unsure of what you're asking. What 2nd part? "How do I find the derivative of ... ?"

Answer 20

ln(xsecx)^(1/2)

Answer 21

What is \[\frac{ d }{ dx }\ln x~?\]

Answer 22

1/x

Answer 23

First I have to verify that I undrstand your math'l expression correctly. Do y ou mean this?\[\frac{ d }{ dx }\ln \sqrt{x*secx)}~?\]

Answer 24

yes

Answer 25

\[y=\sin^{-1} (3x)+\ln \sqrt{x \sec x}\]

Answer 26

We need to take into account / apply properties of logs to simplify this expression. Before we attempt to differentiate ln sqrt(x*sec x), let's rewrite it as follows:\[\ln \sqrt{x*\sec x)}=\ln(x*\sec x)^{1/2}=\frac{ 1 }{ 2 }\ln (x*\sec x)\]

Answer 27

This can be simplified even further. Can you carry out this further simplification?

Answer 28

secx^2 inside the parentheses

Answer 29

?? please explain.

Answer 30

x∗secx=secx^2 ?

Answer 31

No, I'm afraid not. That ' x ' constitutes one function and that sec x constitutes another. You thus have the product of two functions: x * sec x ... and cannot combine them in any way (except perhaps later, when y ou study series representations of functions).

Answer 32

I realized this, yes, in that case I am not sure how to simplify further. My first question is

Answer 33

Your task is to simplify:\[\frac{ 1 }{ 2 }\ln (x*\sec x)\]

Answer 34

how are ln(x∗secx)1/2=1/2ln(x∗secx) equivalent?

Answer 35

through this rule of logarithms:

\[\log a^b = b*\log a\]

Answer 36

\[\log (2x+1)^5=5 \log (2x+1)\]

Answer 37

ok, yes, I remember this rule, it makes sense

Answer 38

\[\log (x*\sec x)^{1/2}=\frac{ 1 }{ 2 }\log (x*\sec x)\]

Answer 39

and that last result is equal to\[\frac{ 1 }{ 2 } [\log (x*\sec x)]=\frac{ 1 }{ 2 }[\log x + \log \sec x]\]

Answer 40

following the rule\[\log a*b = \log a + \log b\]

Answer 41

yes! rule of logs

Answer 42

you can separate

Answer 43

Yes. and after separation, finding the derivative is a lot easier.

Answer 44

\[\frac{ d }{ dx }\frac{ 1 }{ 2 }[\log x + \log \sec x]=\frac{ 1 }{ 2 }[\frac{ 1 }{ x }+\frac{ d }{ dx }\sec x]\]

Answer 45

sum rule I see

Answer 46

right?

Answer 47

Yes. I would like to leave you with this info and to encourage you to finish findng the derivative of the original expression yourself. Although i won't be on OpenStudy all day, I will certainly log on now and then and could then give you further feedback if you need it. Again i encourage you to write down, organize and frequently review formulas such as those we've discussed....this is essential for learning and remembering them.

Answer 48

so then I just add all that together in the end I would think think

Answer 49

since d/dx (f+g)=f'+g'

Answer 50

thanks for the help so far

Answer 51

Any last-minute questions?

Answer 52

By the way, thanks for using Equation Editor. Glad you already have that skill.

Answer 53

So the answer is..

Answer 54

@ganeshie8 @hartnn

Answer 55

\[1/(\sqrt{1-(3x)^{2}}) d/dx +1/2 (\frac{ 1 }{ x }+d/dx \sec x)\]

Answer 56

?

Answer 57

I was hoping you'd summarize our work by typing out what you believe to be the correct answer (the derivative) for this problem. That way, others or I could give you feedback. In the end we all have to develop confidence in judging for ourselves whether or not our results are correct or not.

Answer 58

ah forgot a 3x in there

Answer 59

Your \[1/(\sqrt{1-(3x)^{2}}) d/dx +1/2 (\frac{ 1 }{ x }+d/dx \sec x)\]

Answer 60

I see this as a giant sum rule

Answer 61

with 2 separate parts

Answer 62

is mostly correct. Still needed:

(1) apply the chain rule to that 3x in the first term.
(2) actually find the derivative of sec x; don't just leave it as (d/dx) sec x.

Answer 63

ok

Answer 64

Yes, this problem does involve a "giant sum," and you're correct in applying the sum rule first off.

Answer 65

\[\frac{ d }{ dx }3x = ??\]

Answer 66

3

Answer 67

also, \[\frac{ d }{ dx }\sec x=?\]

Answer 68

secx tan x

Answer 69

Yes. Can you use that information to correct your final answer? Omit the "d/dx" from the first term and omit the (d/dx) sec x from the 2nd term by substituting the appropriate quantities.

Answer 70

ok so it's

Answer 71

\[\frac{ dy }{ dx }=1/(\sqrt{1-(3x)^{2}}) d/dx +1/2 (\frac{ 1 }{ x }+d/dx \sec x)\]

Answer 72

\[3/(\sqrt{1-(3x)^{2}})+\frac{ 1 }{ 2x }+\frac{ \sec x \tan x }{ 2 }\]

Answer 73

Yes, congrats, that's really quite nice.
Look at how we used logs to simplify the differentiation.
Once again: prepare a review sheet including these rules of logs and various differentiation rules. Later you'll want to add integration rules to this list.

Answer 74

ok, will do, thanks again ^_^

Answer 75

Happy to work with you. Keep up the good work!