Help me to find the derivative of a function using log

NIPUL JARIWALA said:
Please help me to find the derivative using Log.
here is my work list. Please spot my mistake.
View attachment 21098
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Your first line of answer is incorrect. You CANNOT reduce "ln" function in that manner. Since you have specific instruction of findinding derivatives by taking logs:

\(\displaystyle \frac{y}{1} \ = \ \frac{e^x + e^{-x}}{e^x - e^{-x}}\)

\(\displaystyle \frac{y +1}{y - 1} \ = \ \frac{(e^x + e^{-x}) + (e^x - e^{-x})}{(e^x + e^{-x}) - (e^x - e^{-x})}\) ...............................property of ratio

\(\displaystyle \frac{y +1}{y - 1} \ = \ \frac{e^x }{e^{-x}}\)

\(\displaystyle \frac{y +1}{y - 1} \ = \ e^{2*x}\)

Now take log and continue....

If I were to do this problem, I would have used "quotient" law of derivatives in first line and see how does "cookie crumble".
 
Yes, i spot the mistake.
I made mistake with log.
Thank you for reply.

And can you tell me, how do you write math equations on this forum?
I write on paper and post its photo.
 
NIPUL JARIWALA said:
Yes, i spot the mistake.
I made mistake with log.
Thank you for reply.

And can you tell me, how do you write math equations on this forum?
I write on paper and post its photo.
Click to expand...
Use LaTex.

If you click on reply to a message using LaTeX, you can see the coding for it and learn by example.

[MATH]\dfrac{e^x}{\pi}.[/MATH]
 
Frankly, I would not use logarithmic differentiation at all for this problem. Just a direct differentiation using the "quotient rule" and the derivative of the exponential is sufficient.
 
The question was asked by my teacher and instructed to use log property to practice it.
And thank you for the reply
 
I like that technique that Subhotosh used. Never saw it before.

[MATH]y = \dfrac{a}{b} \implies \text {both } y + 1 = \dfrac{a}{b} + 1 = \dfrac{a + b}{b}\\ \text {and } y - 1 = \dfrac{a}{b} - 1 = \dfrac{a - b}{b}.\\ \therefore \dfrac{y + 1}{y - 1} = \dfrac{ \dfrac{a + b}{\cancel b}}{\dfrac{a - b}{\cancel b}} = \dfrac{a + b}{a - b}.[/MATH]
 
I think that everybody is making this too hard.

[math] y = \dfrac{e^x+e^{-x}}{e^x-e^{-x}}= \dfrac{\dfrac{e^x+e^{-x}}{2}}{\dfrac{e^x-e^{-x}}{2}}= \dfrac{cosh(x)}{sinh(x)}[/math]
From here you can you logs to make your teacher happy!
 
Jomo said:
I think that everybody is making this too hard.

[math] y = \dfrac{e^x+e^{-x}}{e^x-e^{-x}}= \dfrac{\dfrac{e^x+e^{-x}}{2}}{\dfrac{e^x-e^{-x}}{2}}= \dfrac{cosh(x)}{sinh(x)}[/math]
From here you can you logs to make your teacher happy!
Click to expand...
That's the way i would have done if the specific instruction about taking logs were not present. To continue without logs:

y = coth(x)

y' = - [cosech(x)]2
 
Subhotosh Khan said:
That's the way i would have done if the specific instruction about taking logs were not present. To continue without logs:

y = coth(x)

y' = - [cosech(x)]2
Click to expand...
That's the difference between you and me--I would have done it using hyperbolic functions and then use logs. We have to keep those teachers happy.
 
NIPUL JARIWALA said:
Thank you all of you. You educated me a lot.
Here is the solution which I did when I stared the thread. I was making a mistake with log multiplication rule.
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I would give you full credit for your work but I really believe that the last line should be y' = -[cosech(x)]^2.
Most importantly you learned from this thread and got a very acceptable answer. Good job.
 
I didn't know that the answer was -[cosec(x)]^2
I didn't know about Euler's number with trigonometry.
I learnt it from this thread.
Thank you
 
NIPUL JARIWALA said:
I didn't know that the answer was -[cosec(x)]^2
I didn't know about Euler's number with trigonometry.
I learnt it from this thread.
Thank you
Click to expand...
OK then. Later in calculus you will learn about hyperbolic functions and then you can bring up this problem and look real sharp!
 
NIPUL JARIWALA said:
I didn't know that the answer was -[cosec(x)]^2
I didn't know about Euler's number with trigonometry.
I learnt it from this thread.
Thank you
Click to expand...
No.... It is -[cosech(x)]^2

That h is very important - cannot be ignored.
 
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