Algebra Examples

$\frac{d}{d t} \cdot 6 {(t^{2} + 5 t)}^{- 7}$

Step 1

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

$\frac{d}{d \cdot d} [6 \frac{1}{{(t^{2} + 5 t)}^{7}}]$

Step 2

Simplify the denominator.

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Step 2.1

Factor $t$ out of $t^{2} + 5 t$ .

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Step 2.1.1

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

$\frac{d}{d \cdot d} [6 \frac{1}{{(t \cdot t + 5 t)}^{7}}]$

Step 2.1.2

Factor $t$ out of $5 t$ .

$\frac{d}{d \cdot d} [6 \frac{1}{{(t \cdot t + t \cdot 5)}^{7}}]$

Step 2.1.3

Factor $t$ out of $t \cdot t + t \cdot 5$ .

$\frac{d}{d \cdot d} [6 \frac{1}{{(t (t + 5))}^{7}}]$

Step 2.2

Apply the product rule to $t (t + 5)$ .

$\frac{d}{d \cdot d} [6 \frac{1}{t^{7} {(t + 5)}^{7}}]$

Step 3

Combine $6$ and $\frac{1}{t^{7} {(t + 5)}^{7}}$ .

$\frac{d}{d \cdot d} [\frac{6}{t^{7} {(t + 5)}^{7}}]$

Step 4

Since $\frac{6}{t^{7} {(t + 5)}^{7}}$ is constant with respect to $d$ , the derivative of $\frac{6}{t^{7} {(t + 5)}^{7}}$ with respect to $d$ is $0$ .

$0$