Calculating derivatives for various functions

[mathmari]

Hey! :giggle:

I want to calculate the derivatives of the below functions.

1. $\displaystyle{f(x)=x^n\cdot a^x}$, $\in \mathbb{N}_0, x\in \mathbb{R},a>0$
2. $\displaystyle{f(x)=\log \left [\sqrt{1+\cos^2(x)}\right ]}$,$x\in \mathbb{R}$
3. $\displaystyle{f(x)=\sqrt{e^{\sin \sqrt{x}}}}$, $x>0$
4. $\displaystyle{f(x)=x^p}$, $x>0, p\in \mathbb{R}$
5. $\displaystyle{f(x)=\left (1-\sqrt{2}\sin \left (\frac{x}{2}\right )\right )\cdot \sqrt{1+\tan^2(x)}\cdot \left (1+\sqrt{2}\sin \left (\frac{x}{2}\right )\right )}$, $x\in \left (-\frac{\pi}{2},\frac{\pi}{2}\right )$I have done the following:

Function 1:
The derivative is \begin{equation*}f'(x)=\left (x^n\cdot a^x\right )'=\left (x^n\right )'\cdot a^x+x^n\cdot\left ( a^x\right )'=nx^{n-1}\cdot a^x+x^n\cdot a^x\log a\end{equation*}Function 2:
The derivative is \begin{align*}f'(x)&=\left (\log \left [\sqrt{1+\cos^2(x)}\right ]\right )'=\frac{1}{\sqrt{1+\cos^2(x)}}\cdot \left (\sqrt{1+\cos^2(x)}\right )' \\ & =\frac{1}{\sqrt{1+\cos^2(x)}}\cdot \frac{1}{2\sqrt{1+\cos^2(x)}}\cdot \left (1+\cos^2(x)\right )' =\frac{1}{\sqrt{1+\cos^2(x)}}\cdot \frac{1}{2\sqrt{1+\cos^2(x)}}\cdot 2\cos(x)\cdot \left (\cos (x)\right )' \\ &=\frac{1}{\sqrt{1+\cos^2(x)}}\cdot \frac{1}{2\sqrt{1+\cos^2(x)}}\cdot 2\cos(x)\cdot \left (-\sin (x)\right )=-\frac{\cos (x)\cdot \sin (x)}{1+\cos^2(x)}\end{align*}Function 3:
The derivative is \begin{align*}f'(x)&=\left (\sqrt{e^{\sin \sqrt{x}}}\right )'=\frac{1}{2\sqrt{e^{\sin \sqrt{x}}}}\cdot \left (e^{\sin \sqrt{x}}\right )'=\frac{1}{2\sqrt{e^{\sin \sqrt{x}}}}\cdot e^{\sin \sqrt{x}}\cdot \left (\sin \sqrt{x}\right )'=\frac{1}{2\sqrt{e^{\sin \sqrt{x}}}}\cdot e^{\sin \sqrt{x}}\cdot \cos \sqrt{x} \cdot \left ( \sqrt{x}\right )' \\ & =\frac{1}{2\sqrt{e^{\sin \sqrt{x}}}}\cdot e^{\sin \sqrt{x}}\cdot \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}=\frac{e^{\sin \sqrt{x}}\cdot \cos \sqrt{x}}{4\sqrt{x}\cdot \sqrt{e^{\sin \sqrt{x}}}}=\frac{\left (\sqrt{e^{\sin \sqrt{x}}}\right )^2\cdot \cos \sqrt{x}}{4\sqrt{x}\cdot \sqrt{e^{\sin \sqrt{x}}}}=\frac{\sqrt{e^{\sin \sqrt{x}}}\cdot \cos \sqrt{x}}{4\sqrt{x}} \end{align*}Function 4:
The derivative is \begin{equation*}f'(x)=\left (x^p\right )'=px^{p-1}\end{equation*}Function 5:
The function is equal to \begin{align*}f(x)&=\left (1-\sqrt{2}\sin \left (\frac{x}{2}\right )\right )\cdot \sqrt{1+\tan^2(x)}\cdot \left (1+\sqrt{2}\sin \left (\frac{x}{2}\right )\right )\\ & =\left (1-\sqrt{2}\sin \left (\frac{x}{2}\right )\right )\cdot \left (1+\sqrt{2}\sin \left (\frac{x}{2}\right )\right )\cdot \sqrt{1+\tan^2(x)}\\ & =\left (1^2-\left (\sqrt{2}\sin \left (\frac{x}{2}\right )\right )^2\right )\cdot \sqrt{1+\tan^2(x)}\\ & =\cos (x)\cdot \sqrt{1+\tan^2(x)}\\ & =\cos (x)\cdot \sqrt{1+\frac{\sin^2}{\cos^2}}\\ & =\cos (x)\cdot \sqrt{\frac{\cos^2(x)+\sin^2}{\cos^2}}\\ & =\cos (x)\cdot \sqrt{\frac{1}{\cos^2(x)}}\\ & =\cos (x)\cdot \frac{1}{\cos(x)}\\ & =1\end{align*}
So the derivative is \begin{equation*}f'(x)=\left (1\right )'=0 \end{equation*}
At function 4 it doesn't matter if $p$ is a natural number or a real number, it always like that the derivative, isn't it? For example if $p=\frac{1}{2}$ that formula holds.

:unsure:

 

[I like Serena]

At function 4 it doesn't matter if $p$ is a natural number or a real number, it always like that the derivative, isn't it? For example if $p=\frac{1}{2}$ that formula holds.

Hey mathmari!

What if $p=0$ or $p=1$?

The other derivatives look correct to me. (Nod)

 

[mathmari]

What if $p=0$ or $p=1$?

If $p=0$ then $f(x)=x^0=1$ and $f'(x)=0$. Using the formula $f'(x)=px^{p-1}$ we have $f'(x)=0\cdot x^{0-1}=0$.
If $p=1$ then $f(x)=x^1=x$ and $f'(x)=1$. Using the formula $f'(x)=px^{p-1}$ we have $f'(x)=1\cdot x^{1-1}=x^0=1$.

:unsure:

 

[I like Serena]

If $p=0$ then $f(x)=x^0=1$ and $f'(x)=0$. Using the formula $f'(x)=px^{p-1}$ we have $f'(x)=0\cdot x^{0-1}=0$.
If $p=1$ then $f(x)=x^1=x$ and $f'(x)=1$. Using the formula $f'(x)=px^{p-1}$ we have $f'(x)=1\cdot x^{1-1}=x^0=1$.

These are 'special' at $x=0$. Note that $0\cdot x^{0-1}$ is not defined for $x=0$, but we do have that $(1)'=0$.
Additionally we run into $0^0$, which could be $0$, $1$, undefined, or something else.
So I think we should specify the behavior at $0$.

EDIT: Oh wait. (Wait)
I see now that it is given that $x>0$, in which case these edge cases do not apply.

 

[mathmari]

These are 'special' at $x=0$. Note that $0\cdot x^{0-1}$ is not defined for $x=0$, but we do have that $f'(0)=0$.
Additionally we run into $0^0$, which could be $0$, $1$, undefined, or something else.
So I think we should specify the behavior at $0$.

EDIT: Oh wait. (Wait)
I see now that it is given that $x>0$, in which case these edge cases do not apply.

So for $x>0$ the formula I wrote holds, or not? :unsure:

 

[I like Serena]

So for $x>0$ the formula I wrote holds, or not?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Great! Thank you! (Star)