Find the Derivative - d/dt (1/(t-3))^2

${(\frac{1}{t - 3})}^{2}$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = \frac{1}{t - 3}$ .

To apply the Chain Rule, set $u_{1}$ as $\frac{1}{t - 3}$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d t} [\frac{1}{t - 3}]$

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

$2 u_{1} \frac{d}{d t} [\frac{1}{t - 3}]$

Replace all occurrences of $u_{1}$ with $\frac{1}{t - 3}$ .

$2 \frac{1}{t - 3} \frac{d}{d t} [\frac{1}{t - 3}]$

Combine fractions.

Combine $2$ and $\frac{1}{t - 3}$ .

$\frac{2}{t - 3} \frac{d}{d t} [\frac{1}{t - 3}]$

Rewrite $\frac{1}{t - 3}$ as ${(t - 3)}^{- 1}$ .

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

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = t - 3$ .

To apply the Chain Rule, set $u_{2}$ as $t - 3$ .

$\frac{2}{t - 3} (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d t} [t - 3])$

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

$\frac{2}{t - 3} (- {u_{2}}^{- 2} \frac{d}{d t} [t - 3])$

Replace all occurrences of $u_{2}$ with $t - 3$ .

$\frac{2}{t - 3} (- {(t - 3)}^{- 2} \frac{d}{d t} [t - 3])$

Combine fractions.

Combine $\frac{2}{t - 3}$ and ${(t - 3)}^{- 2}$ .

$- \frac{2 {(t - 3)}^{- 2}}{t - 3} \frac{d}{d t} [t - 3]$

Move ${(t - 3)}^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{2}{(t - 3) {(t - 3)}^{2}} \frac{d}{d t} [t - 3]$

Multiply $t - 3$ by ${(t - 3)}^{2}$ by adding the exponents.

Multiply $t - 3$ by ${(t - 3)}^{2}$ .

Raise $t - 3$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $2$ .

$- \frac{2}{{(t - 3)}^{3}} \frac{d}{d t} [t - 3]$

By the Sum Rule, the derivative of $t - 3$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [- 3]$ .

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

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

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

Since $- 3$ is constant with respect to $t$ , the derivative of $- 3$ with respect to $t$ is $0$ .

$- \frac{2}{{(t - 3)}^{3}} (1 + 0)$

Simplify the expression.

Add $1$ and $0$ .

$- \frac{2}{{(t - 3)}^{3}} \cdot 1$

Multiply $- 1$ by $1$ .

$- \frac{2}{{(t - 3)}^{3}}$

Find the Derivative - d/dt (1/(t-3))^2

Related Questions

Support LCMGCF.com by disabling your adblocker.