Calculus Examples

$y = e^{9 \cos (t) + \ln (t)}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 9 \cos (t) + \ln (t)$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $9 \cos (t) + \ln (t)$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d t} [9 \cos (t) + \ln (t)]$

Step 1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$e^{u} \frac{d}{d t} [9 \cos (t) + \ln (t)]$

Step 1.3

Replace all occurrences of $u$ with $9 \cos (t) + \ln (t)$ .

$e^{9 \cos (t) + \ln (t)} \frac{d}{d t} [9 \cos (t) + \ln (t)]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $9 \cos (t) + \ln (t)$ with respect to $t$ is $\frac{d}{d t} [9 \cos (t)] + \frac{d}{d t} [\ln (t)]$ .

$e^{9 \cos (t) + \ln (t)} (\frac{d}{d t} [9 \cos (t)] + \frac{d}{d t} [\ln (t)])$

Step 2.2

Since $9$ is constant with respect to $t$ , the derivative of $9 \cos (t)$ with respect to $t$ is $9 \frac{d}{d t} [\cos (t)]$ .

$e^{9 \cos (t) + \ln (t)} (9 \frac{d}{d t} [\cos (t)] + \frac{d}{d t} [\ln (t)])$

Step 3

The derivative of $\cos (t)$ with respect to $t$ is $- \sin (t)$ .

$e^{9 \cos (t) + \ln (t)} (9 (- \sin (t)) + \frac{d}{d t} [\ln (t)])$

Step 4

Multiply $- 1$ by $9$ .

$e^{9 \cos (t) + \ln (t)} (- 9 \sin (t) + \frac{d}{d t} [\ln (t)])$

Step 5

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

$e^{9 \cos (t) + \ln (t)} (- 9 \sin (t) + \frac{1}{t})$