Find the Derivative - d/d@VAR s(t) = natural log of (t^2+1)^3

$s (t) = \ln ({(t^{2} + 1)}^{3})$

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

To apply the Chain Rule, set $u_{1}$ as ${(t^{2} + 1)}^{3}$ .

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

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

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

Replace all occurrences of $u_{1}$ with ${(t^{2} + 1)}^{3}$ .

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

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

To apply the Chain Rule, set $u_{2}$ as $t^{2} + 1$ .

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

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

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

Replace all occurrences of $u_{2}$ with $t^{2} + 1$ .

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

Differentiate.

Combine $3$ and $\frac{1}{{(t^{2} + 1)}^{3}}$ .

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

Simplify terms.

Combine ${(t^{2} + 1)}^{2}$ and $\frac{3}{{(t^{2} + 1)}^{3}}$ .

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

Move $3$ to the left of ${(t^{2} + 1)}^{2}$ .

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

Cancel the common factor of ${(t^{2} + 1)}^{2}$ and ${(t^{2} + 1)}^{3}$ .

Factor ${(t^{2} + 1)}^{2}$ out of $3 {(t^{2} + 1)}^{2}$ .

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

Cancel the common factors.

Factor ${(t^{2} + 1)}^{2}$ out of ${(t^{2} + 1)}^{3}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

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

Combine fractions.

Add $2 t$ and $0$ .

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

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

$\frac{2 \cdot 3}{t^{2} + 1} t$

Multiply $2$ by $3$ .

$\frac{6}{t^{2} + 1} t$

Combine $\frac{6}{t^{2} + 1}$ and $t$ .

$\frac{6 t}{t^{2} + 1}$

Find the Derivative - d/d@VAR s(t) = natural log of (t^2+1)^3

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