Question

y = 4x^{2}-\frac{1}{x}
the derivative of
y = 8x - x^{-2}
y = x^{-2}+8x
y = 2x - x^{-2}
y = x^{-1}+8x

Explanation:

Step1: Apply derivative rules

Use the power rule: \((x^n)^\prime = nx^{n - 1}\).
For \(y = 4x^2-\frac{1}{x}=4x^2 - x^{-1}\).
The derivative of \(4x^2\) is \(4\times2x^{2 - 1}=8x\).
The derivative of \(-x^{-1}\) is \(-(-1)x^{-1 - 1}=x^{-2}\).

Step2: Combine derivatives

\(y^\prime=(4x^2 - x^{-1})^\prime=(4x^2)^\prime-(x^{-1})^\prime\).
So \(y^\prime = 8x+x^{-2}\).

Answer:

\(y^\prime=x^{-2}+8x\)