Calculus Examples

$y = 2^{5 x} \cdot 3^{4 x^{2}}$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 2^{5 x}$ and $g (x) = 3^{4 x^{2}}$ .

$2^{5 x} \frac{d}{d x} [3^{4 x^{2}}] + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

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 $g (x) = 4 x^{2}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $4 x^{2}$ .

$2^{5 x} (\frac{d}{d u_{1}} [3^{u_{1}}] \frac{d}{d x} [4 x^{2}]) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $3$ .

$2^{5 x} (3^{u_{1}} \ln (3) \frac{d}{d x} [4 x^{2}]) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 2.3

Replace all occurrences of $u_{1}$ with $4 x^{2}$ .

$2^{5 x} (3^{4 x^{2}} \ln (3) \frac{d}{d x} [4 x^{2}]) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 3

Differentiate using the Constant Multiple Rule.

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Step 3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$2^{5 x} (3^{4 x^{2}} \ln (3) (4 \frac{d}{d x} [x^{2}])) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 3.2

Rewrite $4$ as $2^{2}$ .

$2^{2} \cdot 2^{5 x} (3^{4 x^{2}} \ln (3) (\frac{d}{d x} [x^{2}])) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 4

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

$2^{2 + 5 x} (3^{4 x^{2}} \ln (3) (\frac{d}{d x} [x^{2}])) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 5

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

$2^{2 + 5 x} (3^{4 x^{2}} \ln (3) (2 x)) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 6

Multiply $2^{2 + 5 x}$ by $2$ by adding the exponents.

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Step 6.1

Move $2$ .

$2 \cdot 2^{2 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 6.2

Multiply $2$ by $2^{2 + 5 x}$ .

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Step 6.2.1

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

$2^{1} \cdot 2^{2 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 6.2.2

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

$2^{1 + 2 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 6.3

Add $1$ and $2$ .

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} \frac{d}{d x} [2^{5 x}]$

Step 7

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 $g (x) = 5 x$ .

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Step 7.1

To apply the Chain Rule, set $u_{2}$ as $5 x$ .

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} (\frac{d}{d u_{2}} [2^{u_{2}}] \frac{d}{d x} [5 x])$

Step 7.2

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

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} (2^{u_{2}} \ln (2) \frac{d}{d x} [5 x])$

Step 7.3

Replace all occurrences of $u_{2}$ with $5 x$ .

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} (2^{5 x} \ln (2) \frac{d}{d x} [5 x])$

Step 8

Differentiate.

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Step 8.1

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

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} (2^{5 x} \ln (2) (5 \frac{d}{d x} [x]))$

Step 8.2

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

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} (2^{5 x} \ln (2) (5 \cdot 1))$

Step 8.3

Simplify the expression.

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Step 8.3.1

Multiply $5$ by $1$ .

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} (2^{5 x} \ln (2) \cdot 5)$

Step 8.3.2

Move $5$ to the left of $2^{5 x} \ln (2)$ .

$2^{3 + 5 x} (3^{4 x^{2}} \ln (3) (x)) + 3^{4 x^{2}} (5 \cdot (2^{5 x} \ln (2)))$

Step 9

Simplify.

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Step 9.1

Reorder terms.

$3^{4 x^{2}} \cdot 2^{3 + 5 x} x \ln (3) + 5 \cdot 2^{5 x} \cdot 3^{4 x^{2}} \ln (2)$

Step 9.2

Reorder factors in $3^{4 x^{2}} \cdot 2^{3 + 5 x} x \ln (3) + 5 \cdot 2^{5 x} \cdot 3^{4 x^{2}} \ln (2)$ .

$x \cdot 3^{4 x^{2}} \cdot 2^{3 + 5 x} \ln (3) + 5 \cdot 2^{5 x} \cdot 3^{4 x^{2}} \ln (2)$