Homework Statement

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

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!

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?....

vela
Staff Emeritus
Homework Helper
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.

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.

the original formula is

sin-1(3x)+ln [sqrt(xsecx)]

vela
Staff Emeritus
Homework Helper
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

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

Char. Limit
Gold Member
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.

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.

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

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

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?

Char. Limit
Gold Member
Close. The derivative of sec(x) is sec(x)tan(x), not just tan(x), in the last part.

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!

Char. Limit
Gold Member
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:

$$\left(log\left(f\left(x\right)\right)\right)'=\frac{f'(x)}{f(x)}$$

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

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:

$$\left(log\left(f\left(x\right)\right)\right)'=\frac{f'(x)}{f(x)}$$

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?

Char. Limit
Gold Member
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.