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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 \[e^{e^{e^{e^x}}}\]

Applying chain rule $f'(x)=e^{e^{e^{e^x}}}{e^{e^{e^x}}}{e^{e^x}}e^x$

Find $f''(x)$ where $f(x)=e^{e^{e^{e^x}}}$

Asif Hossain wrote: ↑

Mon Apr 05, 2021 2:48 pm

Find $f''(x)$ where $f(x)=e^{e^{e^{e^x}}}$

Prove that \[\frac{d}{dx}\sin{x}=\cos{x}\]

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 $\lim _{x \to 0} \frac{sin x}{x}=1$]
$=\cos x$

Here goes another easy one
Differentiate $f(x)=x^{(sin(cos(x))}$

Asif Hossain wrote: ↑

Mon Apr 05, 2021 11:08 pm

Here goes another easy one
Differentiate $f(x)=x^{(sin(cos(x))}$

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 $n$.

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?