Calculus Examples

$\frac{d}{d t} \cdot 6 {(t^{2} + 3 t)}^{- 4}$

Step 1

This derivative could not be completed using the chain rule. Mathway will use another method.

Step 2

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

$\frac{d}{d \cdot d} [6 \frac{1}{{(t^{2} + 3 t)}^{4}}]$

Step 3

Simplify the denominator.

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Step 3.1

Factor $t$ out of $t^{2} + 3 t$ .

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Step 3.1.1

Factor $t$ out of $t^{2}$ .

$\frac{d}{d \cdot d} [6 \frac{1}{{(t \cdot t + 3 t)}^{4}}]$

Step 3.1.2

Factor $t$ out of $3 t$ .

$\frac{d}{d \cdot d} [6 \frac{1}{{(t \cdot t + t \cdot 3)}^{4}}]$

Step 3.1.3

Factor $t$ out of $t \cdot t + t \cdot 3$ .

$\frac{d}{d \cdot d} [6 \frac{1}{{(t (t + 3))}^{4}}]$

Step 3.2

Apply the product rule to $t (t + 3)$ .

$\frac{d}{d \cdot d} [6 \frac{1}{t^{4} {(t + 3)}^{4}}]$

Step 4

Combine $6$ and $\frac{1}{t^{4} {(t + 3)}^{4}}$ .

$\frac{d}{d \cdot d} [\frac{6}{t^{4} {(t + 3)}^{4}}]$

Step 5

Since $\frac{6}{t^{4} {(t + 3)}^{4}}$ is constant with respect to $d$ , the derivative of $\frac{6}{t^{4} {(t + 3)}^{4}}$ with respect to $d$ is $0$ .

$0$