Calculus Examples

$- \frac{1}{y^{2}}$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = - 1$ and $g (y) = \frac{1}{y^{2}}$ .

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

Step 2

Differentiate.

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Step 2.1

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

$- \frac{d}{d y} [{(y^{2})}^{- 1}] + \frac{1}{y^{2}} \frac{d}{d y} [- 1]$

Step 2.2

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 2.2.1

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

$- \frac{d}{d y} [y^{2 \cdot - 1}] + \frac{1}{y^{2}} \frac{d}{d y} [- 1]$

Step 2.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3

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

$- (- 2 y^{- 3}) + \frac{1}{y^{2}} \frac{d}{d y} [- 1]$

Step 2.4

Multiply $- 2$ by $- 1$ .

$2 y^{- 3} + \frac{1}{y^{2}} \frac{d}{d y} [- 1]$

Step 2.5

Since $- 1$ is constant with respect to $y$ , the derivative of $- 1$ with respect to $y$ is $0$ .

$2 y^{- 3} + \frac{1}{y^{2}} \cdot 0$

Step 2.6

Simplify the expression.

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Step 2.6.1

Multiply $\frac{1}{y^{2}}$ by $0$ .

$2 y^{- 3} + 0$

Step 2.6.2

Add $2 y^{- 3}$ and $0$ .

$2 y^{- 3}$

Step 3

Simplify.

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Step 3.1

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

$2 \frac{1}{y^{3}}$

Step 3.2

Combine $2$ and $\frac{1}{y^{3}}$ .

$\frac{2}{y^{3}}$