Find dy/dx y=8/(x^2)+4

$y = \frac{8}{x^{2}} + 4$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (\frac{8}{x^{2}} + 4)$

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 $\frac{8}{x^{2}} + 4$ with respect to $x$ is $\frac{d}{d x} [\frac{8}{x^{2}}] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [\frac{8}{x^{2}}] + \frac{d}{d x} [4]$

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

Since $8$ is constant with respect to $x$ , the derivative of $\frac{8}{x^{2}}$ with respect to $x$ is $8 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$8 \frac{d}{d x} [\frac{1}{x^{2}}] + \frac{d}{d x} [4]$

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$8 \frac{d}{d x} [{(x^{2})}^{- 1}] + \frac{d}{d x} [4]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

To apply the Chain Rule, set $u$ as $x^{2}$ .

$8 (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [4]$

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

$8 (- u^{- 2} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [4]$

Replace all occurrences of $u$ with $x^{2}$ .

$8 (- {(x^{2})}^{- 2} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [4]$

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

$8 (- {(x^{2})}^{- 2} (2 x)) + \frac{d}{d x} [4]$

Multiply the exponents in ${(x^{2})}^{- 2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 (- x^{2 \cdot - 2} (2 x)) + \frac{d}{d x} [4]$

Multiply $2$ by $- 2$ .

$8 (- x^{- 4} (2 x)) + \frac{d}{d x} [4]$

Multiply $2$ by $- 1$ .

$8 (- 2 x^{- 4} x) + \frac{d}{d x} [4]$

Raise $x$ to the power of $1$ .

$8 (- 2 (x^{1} x^{- 4})) + \frac{d}{d x} [4]$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$8 (- 2 x^{1 - 4}) + \frac{d}{d x} [4]$

Subtract $4$ from $1$ .

$8 (- 2 x^{- 3}) + \frac{d}{d x} [4]$

Multiply $- 2$ by $8$ .

$- 16 x^{- 3} + \frac{d}{d x} [4]$

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$- 16 x^{- 3} + 0$

Simplify.

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

$- 16 \frac{1}{x^{3}} + 0$

Combine terms.

Combine $- 16$ and $\frac{1}{x^{3}}$ .

$\frac{- 16}{x^{3}} + 0$

Move the negative in front of the fraction.

$- \frac{16}{x^{3}} + 0$

Add $- \frac{16}{x^{3}}$ and $0$ .

$- \frac{16}{x^{3}}$

Step 4

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

$y' = - \frac{16}{x^{3}}$

Step 5

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

$\frac{d y}{d x} = - \frac{16}{x^{3}}$

Find dy/dx y=8/(x^2)+4

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