Differentiation Marathon!

For college and university level advanced Mathematics
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Anindya Biswas
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Differentiation Marathon!

Unread post by Anindya Biswas » Sun Apr 04, 2021 11:29 pm

Thanks for the idea from Asif Hossain. Marathon rules are same. Don't let a problem stay unsolved for more than $2$ days. And post a new derivative problem as soon as one gets solved.
Let's start the journey by a simple one.
$\text{Problem 0 :}$
Find the derivative of the function \[e^{e^{e^{e^x}}}\]
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Asif Hossain
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Re: Differentiation Marathon!

Unread post by Asif Hossain » Mon Apr 05, 2021 12:44 pm

Applying chain rule $f'(x)=e^{e^{e^{e^x}}}{e^{e^{e^x}}}{e^{e^x}}e^x$
Hmm..Hammer...Treat everything as nail

Asif Hossain
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Problem 1

Unread post by Asif Hossain » Mon Apr 05, 2021 2:48 pm

Find $f''(x)$ where $f(x)=e^{e^{e^{e^x}}}$
Hmm..Hammer...Treat everything as nail

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Anindya Biswas
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Solution to Problem 1

Unread post by Anindya Biswas » Mon Apr 05, 2021 5:03 pm

Asif Hossain wrote:
Mon Apr 05, 2021 2:48 pm
Find $f''(x)$ where $f(x)=e^{e^{e^{e^x}}}$
We already know that \[f'(x)=e^{e^{e^{e^x}}}e^{e^{e^x}}e^{e^x}e^x\]
Let's take $\ln$ on both sides.
\[\ln{f'(x)}=e^{e^{e^x}}+e^{e^x}+e^x+x\]
Let's differentiate with respect to $x$ and apply chain-rule.
\[\frac{f''(x)}{f'(x)}=e^{e^{e^x}}e^{e^x}e^x+e^{e^x}e^x+e^x+1\]
\[\Longrightarrow\boxed{f''(x)=e^{e^{e^{e^x}}}e^{e^{e^x}}e^{e^x}e^x\left(e^{e^{e^x}}e^{e^x}e^x+e^{e^x}e^x+e^x+1\right)}\]
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Anindya Biswas
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Problem 2

Unread post by Anindya Biswas » Mon Apr 05, 2021 5:45 pm

Prove that \[\frac{d}{dx}\sin{x}=\cos{x}\]
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mehrab4226
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Re: Differentiation Marathon!

Unread post by Mehrab4226 » Mon Apr 05, 2021 9:41 pm

A formula we are going to use,
$\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2}$
Now,
$\frac{d}{dx}\sin X = \lim _{h \to 0} \frac{\sin (x+h)-\sin x}{h}$
$= \lim _{h \to 0} \frac{2\sin \frac{x+h-x}{2} \cos \frac{2x+h}{2}}{h}$
$= \lim _{h \to 0} \frac{2\sin \frac{h}{2} \cos \frac{2x+h}{2}}{h}$
$= \lim _{h \to 0} \frac{2\sin \frac{h}{2} \cos \frac{2x+h}{2}}{2\frac{h}{2}}$
$= \lim _{\frac{h}{2} \to 0} \frac{2\sin \frac{h}{2}}{2\frac{h}{2}}\times \lim _{h \to 0}\cos (x+\frac{h}{2})$
$=\lim _{\frac{h}{2} \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim _{h \to 0}\cos (x+\frac{h}{2})$
$=1 \times \cos (x+0) $ [since $\lim _{x \to 0} \frac{sin x}{x}=1$]
$=\cos x$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Asif Hossain
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Problem 3

Unread post by Asif Hossain » Mon Apr 05, 2021 11:08 pm

Here goes another easy one
Differentiate $f(x)=x^{(sin(cos(x))}$
Hmm..Hammer...Treat everything as nail

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Anindya Biswas
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Solution to Problem 3

Unread post by Anindya Biswas » Tue Apr 06, 2021 12:14 am

Asif Hossain wrote:
Mon Apr 05, 2021 11:08 pm
Here goes another easy one
Differentiate $f(x)=x^{(sin(cos(x))}$
Let's take $\ln$ on both sides.
\[\ln{f(x)}=\sin{(\cos{x})}\ln{x}\]
\[\Longrightarrow\frac{d}{dx}\ln{f(x)}=\sin{(\cos{x})}\cdot\frac{1}{x}+\cos{(\cos(x))}\cdot(-\sin{x})\cdot\ln{x}\]
\[\Longrightarrow\frac{f'(x)}{f(x)}=\sin{(\cos{x})}\cdot\frac{1}{x}-\cos{(\cos(x))}\cdot\sin{x}\cdot\ln{x}\]
\[\Longrightarrow f'(x)=x^{\sin{(\cos{x})}}\left(\sin{(\cos{x})}\cdot\frac{1}{x}-\cos{(\cos(x))}\cdot\sin{x}\cdot\ln{x}\right)\]
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Anindya Biswas
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Problem 4

Unread post by Anindya Biswas » Tue Apr 06, 2021 1:08 am

Let $f$ be an infinitely differentiable function. Prove that, \[\frac{d^n}{dx^n}\left(e^xf(x)\right)=e^x\sum_{k=0}^{n}{n\choose k}f^{(k)}(x)\] for all nonnegative integer $n$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mehrab4226
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Re: Problem 4

Unread post by Mehrab4226 » Tue Apr 06, 2021 9:37 am

Anindya Biswas wrote:
Tue Apr 06, 2021 1:08 am
Let $f$ be an infinitely differentiable function. Prove that, \[\frac{d^n}{dx^n}\left(e^xf(x)\right)=e^x\sum_{k=0}^{n}{n\choose k}f^{(k)}(x)\] for all nonnegative integer $n$.
For the right hand side,
If k=2, does $f^{(k)}(x)$ mean
$f^{(2)}(x)=f''(x)$ this? or,
$f^{(2)}(x)=f^{2}(x)=f(f(x))$ this?
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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