What is the derivative of ${{\pi }^{x}}$?

Hint: We first try to find the derivative for the general form of ${{a}^{x}}$ where $a$ is constant. We take the logarithm form to find the derivative and then put the value of $a=\pi $ to find the derivative of ${{\pi }^{x}}$.

Complete step by step solution:
we first find the derivative of ${{a}^{x}}$ where $a$ is constant.
We assume that $p={{a}^{x}}$. For differentiation we need to find $\dfrac{dp}{dx}$.
We take logarithms on both sides of the equation and get $\log p=\log \left( {{a}^{x}} \right)$.
We know the identity formula of $\log {{m}^{n}}=n\log m$.
Using the formula, we get $\log p=\log \left( {{a}^{x}} \right)=x\log a$.
As $a$ is constant, the value of $\log a$ also becomes constant.
Now we differentiate both sides of the equation $\log p=x\log a$ with respect to $x$.
We know that differentiation of \[v\left( x \right)=\log x\] is \[{{v}^{'}}\left( x \right)=\dfrac{1}{x}\].
We also apply the formula of \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\].
Therefore, $\dfrac{d}{dx}\left( \log p \right)=\dfrac{d}{dx}\left( x\log a \right)$.
We get $\dfrac{1}{p}\dfrac{dp}{dx}=\log a$. We multiply both sides with $p$ and get $\dfrac{dp}{dx}=p\log a$.
Putting the value of $p$, we get $\dfrac{dp}{dx}={{a}^{x}}\log a$.
Therefore, the differentiation of ${{a}^{x}}$ is ${{a}^{x}}\log a$. We get $\dfrac{dy}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\log a$.
Now to find the differentiation of ${{\pi }^{x}}$, we put $a=\pi $.
So, $\dfrac{dy}{dx}\left( {{\pi }^{x}} \right)={{\pi }^{x}}\log \pi $.
The derivative of ${{\pi }^{x}}$ is ${{\pi }^{x}}\log \pi $.

Note: We find the derivative of ${{e}^{x}}$ using the same formula where we take $a=e$.
Therefore, $\dfrac{dy}{dx}\left( {{e}^{x}} \right)={{e}^{x}}\log e$. The base of the logarithm is also $e$ and therefore, the value of
$\dfrac{dy}{dx}\left( {{e}^{x}} \right)={{e}^{x}}\log e={{e}^{x}}$ as we know ${{\log }_{m}}m=1$.