Calculus Examples

$y = x^{4} + 8$

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (x^{4} + 8)$

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

$y'$

Differentiate the right side of the equation.

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By the Sum Rule, the derivative of $x^{4} + 8$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [8]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [8]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$4 x^{3} + \frac{d}{d x} [8]$

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$4 x^{3} + 0$

Add $4 x^{3}$ and $0$ .

$4 x^{3}$

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

$y' = 4 x^{3}$

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

$\frac{d y}{d x} = 4 x^{3}$

Set $\frac{d y}{d x} = 0$ then solve for $x$ in terms of $y$ .

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Divide each term by $4$ and simplify.

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Divide each term in $4 x^{3} = 0$ by $4$ .

$\frac{4 x^{3}}{4} = \frac{0}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 x^{3}}{4} = \frac{0}{4}$

Divide $x^{3}$ by $1$ .

$x^{3} = \frac{0}{4}$

Divide $0$ by $4$ .

$x^{3} = 0$

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

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Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Simplify ${(0)}^{4} + 8$ .

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Raising $0$ to any positive power yields $0$ .

$y = 0 + 8$

Add $0$ and $8$ .

$y = 8$

Find the points where $\frac{d y}{d x} = 0$ .

$(0, 8)$

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