Find the Derivative - d/dt d/(dt)*6(t^2+5t)^-7

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

Step 1

Since $6$ is constant with respect to $t$ , the derivative of $6 {(t^{2} + 5 t)}^{- 7}$ with respect to $t$ is $6 \frac{d}{d t} [{(t^{2} + 5 t)}^{- 7}]$ .

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 7}$ and $g (t) = t^{2} + 5 t$ .

To apply the Chain Rule, set $u$ as $t^{2} + 5 t$ .

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

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 7$ .

$6 (- 7 u^{- 8} \frac{d}{d t} [t^{2} + 5 t])$

Replace all occurrences of $u$ with $t^{2} + 5 t$ .

$6 (- 7 {(t^{2} + 5 t)}^{- 8} \frac{d}{d t} [t^{2} + 5 t])$

Step 3

Differentiate.

Multiply $- 7$ by $6$ .

$- 42 ({(t^{2} + 5 t)}^{- 8} \frac{d}{d t} [t^{2} + 5 t])$

By the Sum Rule, the derivative of $t^{2} + 5 t$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [5 t]$ .

$- 42 {(t^{2} + 5 t)}^{- 8} (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [5 t])$

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

$- 42 {(t^{2} + 5 t)}^{- 8} (2 t + \frac{d}{d t} [5 t])$

Since $5$ is constant with respect to $t$ , the derivative of $5 t$ with respect to $t$ is $5 \frac{d}{d t} [t]$ .

$- 42 {(t^{2} + 5 t)}^{- 8} (2 t + 5 \frac{d}{d t} [t])$

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

$- 42 {(t^{2} + 5 t)}^{- 8} (2 t + 5 \cdot 1)$

Multiply $5$ by $1$ .

$- 42 {(t^{2} + 5 t)}^{- 8} (2 t + 5)$

Step 4

Simplify.

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

$- 42 \frac{1}{{(t^{2} + 5 t)}^{8}} (2 t + 5)$

Combine terms.

Combine $- 42$ and $\frac{1}{{(t^{2} + 5 t)}^{8}}$ .

$\frac{- 42}{{(t^{2} + 5 t)}^{8}} (2 t + 5)$

Move the negative in front of the fraction.

$- \frac{42}{{(t^{2} + 5 t)}^{8}} (2 t + 5)$

Reorder the factors of $- \frac{42}{{(t^{2} + 5 t)}^{8}} (2 t + 5)$ .

$- (2 t + 5) \frac{42}{{(t^{2} + 5 t)}^{8}}$

Apply the distributive property.

$(- (2 t) - 1 \cdot 5) \frac{42}{{(t^{2} + 5 t)}^{8}}$

Multiply $2$ by $- 1$ .

$(- 2 t - 1 \cdot 5) \frac{42}{{(t^{2} + 5 t)}^{8}}$

Multiply $- 1$ by $5$ .

$(- 2 t - 5) \frac{42}{{(t^{2} + 5 t)}^{8}}$

Simplify the denominator.

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

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

$(- 2 t - 5) \frac{42}{{(t \cdot t + 5 t)}^{8}}$

Factor $t$ out of $5 t$ .

$(- 2 t - 5) \frac{42}{{(t \cdot t + t \cdot 5)}^{8}}$

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

$(- 2 t - 5) \frac{42}{{(t (t + 5))}^{8}}$

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

$(- 2 t - 5) \frac{42}{t^{8} {(t + 5)}^{8}}$

Multiply $- 2 t - 5$ by $\frac{42}{t^{8} {(t + 5)}^{8}}$ .

$\frac{(- 2 t - 5) \cdot 42}{t^{8} {(t + 5)}^{8}}$

Move $42$ to the left of $- 2 t - 5$ .

$\frac{42 \cdot (- 2 t - 5)}{t^{8} {(t + 5)}^{8}}$

Factor $- 1$ out of $- 2 t$ .

$\frac{42 (- (2 t) - 5)}{t^{8} {(t + 5)}^{8}}$

Rewrite $- 5$ as $- 1 (5)$ .

$\frac{42 (- (2 t) - 1 (5))}{t^{8} {(t + 5)}^{8}}$

Factor $- 1$ out of $- (2 t) - 1 (5)$ .

$\frac{42 (- (2 t + 5))}{t^{8} {(t + 5)}^{8}}$

Rewrite $- (2 t + 5)$ as $- 1 (2 t + 5)$ .

$\frac{42 (- 1 (2 t + 5))}{t^{8} {(t + 5)}^{8}}$

Move the negative in front of the fraction.

$- \frac{42 (2 t + 5)}{t^{8} {(t + 5)}^{8}}$

Find the Derivative - d/dt d/(dt)*6(t^2+5t)^-7

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