Is This the Correct Derivative for sin^(-1)(3x) + ln(sqrt(x) sqrt(sec x))?

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Homework Help Overview

The discussion revolves around finding the derivative of the function sin-1(3x) + ln(sqrt(x) sqrt(sec x)). Participants are examining the notation and the application of logarithmic properties in the context of differentiation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the notation used in the original problem, particularly whether sin-1(3x) refers to arcsin(3x) or 1/sin(3x). There are discussions about the application of logarithmic properties to simplify the second term before differentiation.

Discussion Status

Some participants have provided guidance on how to approach the differentiation, suggesting the use of logarithmic properties and clarifying the meaning of the terms involved. There is an ongoing exploration of different interpretations and calculations, with no explicit consensus reached on the final answer.

Contextual Notes

There are indications of ambiguity in the original notation, particularly regarding the application of the natural logarithm and the interpretation of the first term. Participants are also discussing potential errors in derivative calculations and the reliability of online tools for verification.

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Homework Statement



sin-13x+ln sqrt(x) sqrt(sec x)

Homework Equations


The Attempt at a Solution


I have a very long solution, which is hard to type in. My final answer is:
-3cos3x/sin23x + (1+xtanx)/2x

I don't check it with a derivative calculator because usually it makes mistakes. I ask someone to solve the problem and see if this is the right answer.
Thank you!
 
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This is very interesting...My answer is completely different from the given on the website. I also tweaked a formula, however, it doesn't help, the answer is still far off mine.
Can anyone solve the derivative please by hand?...
 
It would help if you clarified exactly what the function is. Your notation is a bit ambiguous. Besides the question about the log that Raskolnikov raised, is the first term the arcsin or is it the reciprocal of sin? That notation usually means arcsin 3x, but you've interpreted it as 1/sin 3x.
 
I love WolframAlpha. There's an option to show the solution for the derivation. Just click the option "show steps" located on the upper-right corner of the Derivative box where it displays the answer.
 
vela said:
It would help if you clarified exactly what the function is. Your notation is a bit ambiguous. Besides the question about the log that Raskolnikov raised, is the first term the arcsin or is it the reciprocal of sin? That notation usually means arcsin 3x, but you've interpreted it as 1/sin 3x.

My bad.

the original formula is

sin-1(3x)+ln [sqrt(xsecx)]
 
The first term means arcsin 3x, not 1/(sin 3x). So try differentiating that.

Differentiating the second term is a bit simpler if you use a properties of the log to rewrite it as

[tex]\log \sqrt{x \sec x} = \log (x \sec x)^{1/2} = \frac{1}{2} \log (x \sec x) = \frac{1}{2} (\log x + \log \sec x)[/tex]
 
Raskolnikov said:
I love WolframAlpha. There's an option to show the solution for the derivation. Just click the option "show steps" located on the upper-right corner of the Derivative box where it displays the answer.

Sometimes W|A integrates oddly though... I once saw it do four u-subs to the same variable chain, when a simple parts integral would suffice.
 
Char. Limit said:
Sometimes W|A integrates oddly though... I once saw it do four u-subs to the same variable chain, when a simple parts integral would suffice.

true true...probably some programming miscalculation. it's still a nice site on a whole though. and wolfram's mathematica site is pretty useful too.
 
  • #10
vela said:
The first term means arcsin 3x, not 1/(sin 3x). So try differentiating that.

Differentiating the second term is a bit simpler if you use a properties of the log to rewrite it as

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

Based on your suggestion, I got:

arcsin 3x+1/2(logx+logsecx)
3/(sqrt(1-(3x)2)+1/(2x)+1/(2secx)*tanx

Am I right?
 
  • #11
Close. The derivative of sec(x) is sec(x)tan(x), not just tan(x), in the last part.
 
  • #12
Char. Limit said:
Close. The derivative of sec(x) is sec(x)tan(x), not just tan(x), in the last part.

OK, so the last part of my equation is

tanx/2secxtanx=1/2secx

?
Thank you!
 
  • #13
3songs said:
OK, so the last part of my equation is

tanx/2secxtanx=1/2secx

?
Thank you!

Not quite. Originally, you wrote 1/sec(x) * tan(x). This was close, but it needs to be (1/sec(x))(sec(x)tan(x)).

Consider this little fact:

[tex]\left(log\left(f\left(x\right)\right)\right)'=\frac{f'(x)}{f(x)}[/tex]

So, since f(x)=sec(x), f'(x)=sec(x)tan(x).
 
  • #14
Char. Limit said:
Not quite. Originally, you wrote 1/sec(x) * tan(x). This was close, but it needs to be (1/sec(x))(sec(x)tan(x)).

Consider this little fact:

[tex]\left(log\left(f\left(x\right)\right)\right)'=\frac{f'(x)}{f(x)}[/tex]

So, since f(x)=sec(x), f'(x)=sec(x)tan(x).

How silly I am.
After I make these changes, I have:

3/(sqrt(1-(3x)2) +1/(2x) +tanx/2

Is this the final answer?
 
  • #15
3songs said:
How silly I am.
After I make these changes, I have:

3/(sqrt(1-(3x)2) +1/(2x) +tanx/2

Is this the final answer?

Looks good to me.
 

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