Calculus Examples

$y = - 7 x (8^{- 2 x})$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (- 7 x \cdot 8^{- 2 x})$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7 x \cdot 8^{- 2 x}$ with respect to $x$ is $- 7 \frac{d}{d x} [x \cdot 8^{- 2 x}]$ .

$- 7 \frac{d}{d x} [x \cdot 8^{- 2 x}]$

Step 3.2

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 $f (x) = x$ and $g (x) = 8^{- 2 x}$ .

$- 7 (x \frac{d}{d x} [8^{- 2 x}] + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 8^{x}$ and $g (x) = - 2 x$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $- 2 x$ .

$- 7 (x (\frac{d}{d u} [8^{u}] \frac{d}{d x} [- 2 x]) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $8$ .

$- 7 (x (8^{u} \ln (8) \frac{d}{d x} [- 2 x]) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.3.3

Replace all occurrences of $u$ with $- 2 x$ .

$- 7 (x (8^{- 2 x} \ln (8) \frac{d}{d x} [- 2 x]) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4

Differentiate.

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Step 3.4.1

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

$- 7 (x (8^{- 2 x} \ln (8) (- 2 \frac{d}{d x} [x])) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4.2

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

$- 7 (x (8^{- 2 x} \ln (8) (- 2 \cdot 1)) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4.3

Simplify with factoring out.

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Step 3.4.3.1

Multiply $- 2$ by $1$ .

$- 7 (x (8^{- 2 x} \ln (8) \cdot - 2) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4.3.2

Move $- 2$ to the left of $8^{- 2 x} \ln (8)$ .

$- 7 (x (- 2 \cdot (8^{- 2 x} \ln (8))) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4.3.3

Factor out negative.

$- 7 (x (- (2 \cdot 8^{- 2 x}) \ln (8)) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4.3.4

Simplify the expression.

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Step 3.4.3.4.1

Rewrite $8$ as $2^{3}$ .

$- 7 (x (- (2 \cdot {(2^{3})}^{- 2 x}) \ln (8)) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4.3.4.2

Multiply the exponents in ${(2^{3})}^{- 2 x}$ .

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Step 3.4.3.4.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- 7 (x (- (2 \cdot 2^{3 (- 2 x)}) \ln (8)) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.4.3.4.2.2

Multiply $- 2$ by $3$ .

$- 7 (x (- (2 \cdot 2^{- 6 x}) \ln (8)) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.5

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

$- 7 (x (- 2^{1 - 6 x} \ln (8)) + 8^{- 2 x} \frac{d}{d x} [x])$

Step 3.6

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

$- 7 (x (- 2^{1 - 6 x} \ln (8)) + 8^{- 2 x} \cdot 1)$

Step 3.7

Multiply $8^{- 2 x}$ by $1$ .

$- 7 (x (- 2^{1 - 6 x} \ln (8)) + 8^{- 2 x})$

Step 3.8

Simplify.

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Step 3.8.1

Apply the distributive property.

$- 7 (x (- 2^{1 - 6 x} \ln (8))) - 7 \cdot 8^{- 2 x}$

Step 3.8.2

Multiply $- 1$ by $- 7$ .

$7 x (2^{1 - 6 x} \ln (8)) - 7 \cdot 8^{- 2 x}$

Step 3.8.3

Reorder terms.

$7 \cdot 2^{1 - 6 x} x \ln (8) - 7 \cdot 8^{- 2 x}$

Step 3.8.4

Reorder factors in $7 \cdot 2^{1 - 6 x} x \ln (8) - 7 \cdot 8^{- 2 x}$ .

$7 x \cdot 2^{1 - 6 x} \ln (8) - 7 \cdot 8^{- 2 x}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = 7 x \cdot (2^{1 - 6 x} \ln (8)) - 7 \cdot 8^{- 2 x}$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = 7 x \cdot (2^{1 - 6 x} \ln (8)) - 7 \cdot 8^{- 2 x}$