Question

ex #2: f(x)=\sqrt4{x}

Explanation:

Since the problem statement is incomplete (it just shows the function \( f(x)=\sqrt[4]{x} \) without a specific question like finding the derivative, domain, etc.), I'll assume a common task like finding the derivative.

Step1: Rewrite the function

Rewrite \( \sqrt[4]{x} \) as a power function. Recall that \( \sqrt[n]{x}=x^{\frac{1}{n}} \), so \( \sqrt[4]{x}=x^{\frac{1}{4}} \).

Step2: Apply the power rule

The power rule for differentiation is \( \frac{d}{dx}(x^n)=nx^{n - 1} \). Here, \( n=\frac{1}{4} \), so:
\( f^\prime(x)=\frac{1}{4}x^{\frac{1}{4}-1}=\frac{1}{4}x^{-\frac{3}{4}} \)
We can also rewrite \( x^{-\frac{3}{4}} \) as \( \frac{1}{x^{\frac{3}{4}}}=\frac{1}{\sqrt[4]{x^3}} \), so \( f^\prime(x)=\frac{1}{4\sqrt[4]{x^3}} \) (with the domain \( x>0 \) since the original function \( \sqrt[4]{x} \) is defined for \( x\geq0 \), but the derivative at \( x = 0 \) is undefined as the exponent of \( x \) in the derivative is negative).

Answer:

If finding the derivative, the derivative of \( f(x)=\sqrt[4]{x} \) is \( f^\prime(x)=\frac{1}{4}x^{-\frac{3}{4}} \) or \( \frac{1}{4\sqrt[4]{x^3}} \) (for \( x>0 \))