Algebra Examples

$t = z^{2} + \frac{6}{z} + 3 z^{- 2}$

Step 1

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

$\frac{d}{d t} [z^{2} + \frac{6}{z} + 3 \frac{1}{z^{2}}]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $z^{2} + \frac{6}{z} + 3 \frac{1}{z^{2}}$ with respect to $t$ is $\frac{d}{d t} [z^{2}] + \frac{d}{d t} [\frac{6}{z}] + \frac{d}{d t} [3 \frac{1}{z^{2}}]$ .

$\frac{d}{d t} [z^{2}] + \frac{d}{d t} [\frac{6}{z}] + \frac{d}{d t} [3 \frac{1}{z^{2}}]$

Step 2.2

Since $z^{2}$ is constant with respect to $t$ , the derivative of $z^{2}$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [\frac{6}{z}] + \frac{d}{d t} [3 \frac{1}{z^{2}}]$

Step 2.3

Since $\frac{6}{z}$ is constant with respect to $t$ , the derivative of $\frac{6}{z}$ with respect to $t$ is $0$ .

$0 + 0 + \frac{d}{d t} [3 \frac{1}{z^{2}}]$

Step 3

Evaluate $\frac{d}{d t} [3 \frac{1}{z^{2}}]$ .

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Step 3.1

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

$0 + 0 + \frac{d}{d t} [\frac{3}{z^{2}}]$

Step 3.2

Since $\frac{3}{z^{2}}$ is constant with respect to $t$ , the derivative of $\frac{3}{z^{2}}$ with respect to $t$ is $0$ .

$0 + 0 + 0$

Step 4

Combine terms.

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Step 4.1

Add $0$ and $0$ .

$0 + 0$

Step 4.2

Add $0$ and $0$ .

$0$