Calculus Examples

$y = 2^{3^{x^{2}}}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 2^{x}$ and $g (x) = 3^{x^{2}}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $3^{x^{2}}$ .

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

Step 1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $2$ .

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

Step 1.3

Replace all occurrences of $u_{1}$ with $3^{x^{2}}$ .

$2^{3^{x^{2}}} \ln (2) \frac{d}{d x} [3^{x^{2}}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 3^{x}$ and $g (x) = x^{2}$ .

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as $x^{2}$ .

$2^{3^{x^{2}}} \ln (2) (\frac{d}{d u_{2}} [3^{u_{2}}] \frac{d}{d x} [x^{2}])$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $3$ .

$2^{3^{x^{2}}} \ln (2) (3^{u_{2}} \ln (3) \frac{d}{d x} [x^{2}])$

Step 2.3

Replace all occurrences of $u_{2}$ with $x^{2}$ .

$2^{3^{x^{2}}} \ln (2) (3^{x^{2}} \ln (3) \frac{d}{d x} [x^{2}])$

Step 3

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

$2^{3^{x^{2}}} \ln (2) (3^{x^{2}} \ln (3) (2 x))$

Step 4

Raise $2$ to the power of $1$ .

$2^{1} \cdot 2^{3^{x^{2}}} \ln (2) (3^{x^{2}} \ln (3) (x))$

Step 5

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

$2^{1 + 3^{x^{2}}} \ln (2) (3^{x^{2}} \ln (3) (x))$

Step 6

Simplify.

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Step 6.1

Reorder the factors of $2^{1 + 3^{x^{2}}} \ln (2) (3^{x^{2}} \ln (3) (x))$ .

$2^{1 + 3^{x^{2}}} \cdot 3^{x^{2}} x \ln (2) \ln (3)$

Step 6.2

Reorder factors in $2^{1 + 3^{x^{2}}} \cdot 3^{x^{2}} x \ln (2) \ln (3)$ .

$x \cdot 2^{1 + 3^{x^{2}}} \cdot 3^{x^{2}} \ln (2) \ln (3)$