Find dy/dx y=8x^2+2/x

$y = 8 x^{2} + \frac{2}{x}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (8 x^{2} + \frac{2}{x})$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

By the Sum Rule, the derivative of $8 x^{2} + \frac{2}{x}$ with respect to $x$ is $\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [\frac{2}{x}]$ .

$\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [\frac{2}{x}]$

Evaluate $\frac{d}{d x} [8 x^{2}]$ .

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{2}$ with respect to $x$ is $8 \frac{d}{d x} [x^{2}]$ .

$8 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [\frac{2}{x}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$8 (2 x) + \frac{d}{d x} [\frac{2}{x}]$

Multiply $2$ by $8$ .

$16 x + \frac{d}{d x} [\frac{2}{x}]$

Evaluate $\frac{d}{d x} [\frac{2}{x}]$ .

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2}{x}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{1}{x}]$ .

$16 x + 2 \frac{d}{d x} [\frac{1}{x}]$

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$16 x + 2 \frac{d}{d x} [x^{- 1}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 1$ .

$16 x + 2 (- x^{- 2})$

Multiply $- 1$ by $2$ .

$16 x - 2 x^{- 2}$

Simplify.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$16 x - 2 \frac{1}{x^{2}}$

Combine terms.

Combine $- 2$ and $\frac{1}{x^{2}}$ .

$16 x + \frac{- 2}{x^{2}}$

Move the negative in front of the fraction.

$16 x - \frac{2}{x^{2}}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = 16 x - \frac{2}{x^{2}}$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = 16 x - \frac{2}{x^{2}}$