Solveeit Logo
Login

Question

Question: How do you find the derivative of \[f\left( x \right)={{x}^{\pi }}?\]...

How do you find the derivative of f(x)=xπ?f\left( x \right)={{x}^{\pi }}?

Explanation

Solution

We are given a function as f(x)=xπf\left( x \right)={{x}^{\pi }} and we are asked to find the derivative of it. First, we will learn what derivative is and then we will use the formula d(xn)dx=nxn−1,\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}, where x is a variable and n is any constant. Then we will also learn that integration is just the opposite of differentiation. So, we will use integration to check our solution whether it is correct or not.

Complete step-by-step solution:
We are given a function as f(x)=xπ.f\left( x \right)={{x}^{\pi }}. Before we move forward we will learn about the derivative. Derivative or differentiation of any function means the rate of change of a function with respect to its variable. This is equivalent to finding the slope of the tangent line of the function at a point. Now, we know the different function has a different derivation, the function of the form xn{{x}^{n}} where x is variable and n is constant. If the derivative is given as nxn−1n{{x}^{n-1}} that is d(xn)dx=nxn−1.\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}. We can see that our function is xπ,{{x}^{\pi }}, so clearly, it is of the type xn.{{x}^{n}}. So, we will use the above defined formula to find its derivative. For this, our n will be π.\pi .
d(xn)dx=nxn−1\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}
Here, n=π,n=\pi , so we get,
⇒d(xπ)dx=πxπ−1\Rightarrow \dfrac{d\left( {{x}^{\pi }} \right)}{dx}=\pi {{x}^{\pi -1}}
So, we get the derivative of xπ{{x}^{\pi }} as πxπ−1.\pi {{x}^{\pi -1}}.

Note: Remember that the derivative is defined as the rate of change of a function with variable means we have divided the whole part into smaller parts. While integration is a way of adding slices to find the whole. So, clearly, it means that integration is just the opposite of derivation. So, we integrate our solution, if it came as original then it is the correct solution. Now, we know ∫xndx=xn+1n+1.\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}.} Now we have πxπ−1,\pi {{x}^{\pi -1}}, so we get,
∫πxπ−1dx=π∫xπ−1dx\int{\pi {{x}^{\pi -1}}dx}=\pi \int{{{x}^{\pi -1}}dx}
Here, n=π−1,n=\pi -1, so we get,
=π[xπ−1+1π−1+1]=\pi \left[ \dfrac{{{x}^{\pi -1+1}}}{\pi -1+1} \right]
On simplifying, we get,
=π[xππ]=\pi \left[ \dfrac{{{x}^{\pi }}}{\pi } \right]
Cancelling π\pi we get,
=xπ={{x}^{\pi }}
This is the original function. Hence our solution is correct.