Calculus Examples

$\cos (π x)$

Write $\cos (π x)$ as a function.

$f (x) = \cos (π x)$

Find the first derivative of the function.

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

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

$\frac{d}{d u} [\cos (u)] \frac{d}{d x} [π x]$

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

$- \sin (u) \frac{d}{d x} [π x]$

Replace all occurrences of $u$ with $π x$ .

$- \sin (π x) \frac{d}{d x} [π x]$

Differentiate.

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

$- \sin (π x) (π \frac{d}{d x} [x])$

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

$- \sin (π x) (π \cdot 1)$

Simplify the expression.

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Multiply $π$ by $1$ .

$- \sin (π x) π$

Reorder the factors of $- \sin (π x) π$ .

$- π \sin (π x)$

Find the second derivative of the function.

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

$f'' (x) = - π \frac{d}{d x} \sin (π x)$

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

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

$f'' (x) = - π (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (π x))$

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

$f'' (x) = - π (\cos (u) \frac{d}{d x} (π x))$

Replace all occurrences of $u$ with $π x$ .

$f'' (x) = - π (\cos (π x) \frac{d}{d x} (π x))$

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

$f'' (x) = - π \cos (π x) (π \frac{d}{d x} (x))$

Raise $π$ to the power of $1$ .

$f'' (x) = - π π \cos (π x) \frac{d}{d x} x$

Raise $π$ to the power of $1$ .

$f'' (x) = - π π \cos (π x) \frac{d}{d x} x$

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

$f'' (x) = - π^{1 + 1} \cos (π x) \frac{d}{d x} x$

Add $1$ and $1$ .

$f'' (x) = - π^{2} \cos (π x) \frac{d}{d x} x$

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

$f'' (x) = - π^{2} \cos (π x) \cdot 1$

Multiply $- 1$ by $1$ .

$f'' (x) = - π^{2} \cos (π x)$

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- π \sin (π x) = 0$

Divide each term in $- π \sin (π x) = 0$ by $- π$ and simplify.

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Divide each term in $- π \sin (π x) = 0$ by $- π$ .

$\frac{- π \sin (π x)}{- π} = \frac{0}{- π}$

Simplify the left side.

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Dividing two negative values results in a positive value.

$\frac{π \sin (π x)}{π} = \frac{0}{- π}$

Cancel the common factor of $π$ .

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Cancel the common factor.

$\frac{π \sin (π x)}{π} = \frac{0}{- π}$

Divide $\sin (π x)$ by $1$ .

$\sin (π x) = \frac{0}{- π}$

Simplify the right side.

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Divide $0$ by $- π$ .

$\sin (π x) = 0$

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$π x = \arcsin (0)$

Simplify the right side.

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The exact value of $\arcsin (0)$ is $0$ .

$π x = 0$

Divide each term in $π x = 0$ by $π$ and simplify.

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Divide each term in $π x = 0$ by $π$ .

$\frac{π x}{π} = \frac{0}{π}$

Simplify the left side.

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Cancel the common factor of $π$ .

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Cancel the common factor.

$\frac{π x}{π} = \frac{0}{π}$

Divide $x$ by $1$ .

$x = \frac{0}{π}$

Simplify the right side.

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Divide $0$ by $π$ .

$x = 0$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$π x = π - 0$

Solve for $x$ .

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Simplify.

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Multiply $- 1$ by $0$ .

$π x = π + 0$

Add $π$ and $0$ .

$π x = π$

Divide each term in $π x = π$ by $π$ and simplify.

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Divide each term in $π x = π$ by $π$ .

$\frac{π x}{π} = \frac{π}{π}$

Simplify the left side.

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Cancel the common factor of $π$ .

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Cancel the common factor.

$\frac{π x}{π} = \frac{π}{π}$

Divide $x$ by $1$ .

$x = \frac{π}{π}$

Simplify the right side.

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Cancel the common factor of $π$ .

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Cancel the common factor.

$x = \frac{π}{π}$

Rewrite the expression.

$x = 1$

The solution to the equation $π x = 0$ .

$x = 0, 1$

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- π^{2} \cos (π (0))$

Evaluate the second derivative.

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Multiply $π$ by $0$ .

$- π^{2} \cos (0)$

The exact value of $\cos (0)$ is $1$ .

$- π^{2} \cdot 1$

Multiply $- 1$ by $1$ .

$- π^{2}$

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Find the y-value when $x = 0$ .

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Replace the variable $x$ with $0$ in the expression.

$f (0) = \cos (π (0))$

Simplify the result.

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Multiply $π$ by $0$ .

$f (0) = \cos (0)$

The exact value of $\cos (0)$ is $1$ .

$f (0) = 1$

The final answer is $1$ .

$y = 1$

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- π^{2} \cos (π (1))$

Evaluate the second derivative.

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Multiply $π$ by $1$ .

$- π^{2} \cos (π)$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$- π^{2} (- \cos (0))$

The exact value of $\cos (0)$ is $1$ .

$- π^{2} (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$- π^{2} \cdot - 1$

Multiply $- π^{2} \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 π^{2}$

Multiply $π^{2}$ by $1$ .

$π^{2}$

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Find the y-value when $x = 1$ .

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Replace the variable $x$ with $1$ in the expression.

$f (1) = \cos (π (1))$

Simplify the result.

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Multiply $π$ by $1$ .

$f (1) = \cos (π)$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$f (1) = - \cos (0)$

The exact value of $\cos (0)$ is $1$ .

$f (1) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f (1) = - 1$

The final answer is $- 1$ .

$y = - 1$

These are the local extrema for $f (x) = \cos (π x)$ .

$(0, 1)$ is a local maxima

$(1, - 1)$ is a local minima