Calculus Examples

$y = {(x + 3)}^{2}$ , $(1, 16)$

The slope of the tangent line is the derivative of the expression.

$m$ $=$ The derivative of $y = {(x + 3)}^{2}$

Consider the limit definition of the derivative.

$f' (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Find the components of the definition.

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Evaluate the function at $x = x + h$ .

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Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = {((x + h) + 3)}^{2}$

Simplify the result.

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Simplify the expression.

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Remove parentheses around $x + h$ .

$f (x + h) = {(x + h + 3)}^{2}$

Rewrite ${(x + h + 3)}^{2}$ as $(x + h + 3) (x + h + 3)$ .

$f (x + h) = (x + h + 3) (x + h + 3)$

Expand $(x + h + 3) (x + h + 3)$ by multiplying each term in the first expression by each term in the second expression.

$f (x + h) = x x + x h + x \cdot 3 + (h x + h h + h \cdot 3) + (3 x + 3 h + 3 \cdot 3)$

Simplify terms.

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Remove unnecessary parentheses.

$f (x + h) = x x + x h + x \cdot 3 + h x + h h + h \cdot 3 + 3 x + 3 h + 3 \cdot 3$

Simplify each term.

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

$f (x + h) = x^{1 + 1} + x h + x \cdot 3 + h x + h h + h \cdot 3 + 3 x + 3 h + 3 \cdot 3$

Add $1$ and $1$ to get $2$ .

$f (x + h) = x^{2} + x h + x \cdot 3 + h x + h h + h \cdot 3 + 3 x + 3 h + 3 \cdot 3$

Move $3$ to the left of the expression $x \cdot 3$ .

$f (x + h) = x^{2} + x h + 3 \cdot x + h x + h h + h \cdot 3 + 3 x + 3 h + 3 \cdot 3$

Multiply $3$ by $x$ to get $3 x$ .

$f (x + h) = x^{2} + x h + 3 x + h x + h h + h \cdot 3 + 3 x + 3 h + 3 \cdot 3$

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

$f (x + h) = x^{2} + x h + 3 x + h x + h^{1 + 1} + h \cdot 3 + 3 x + 3 h + 3 \cdot 3$

Add $1$ and $1$ to get $2$ .

$f (x + h) = x^{2} + x h + 3 x + h x + h^{2} + h \cdot 3 + 3 x + 3 h + 3 \cdot 3$

Move $3$ to the left of the expression $h \cdot 3$ .

$f (x + h) = x^{2} + x h + 3 x + h x + h^{2} + 3 \cdot h + 3 x + 3 h + 3 \cdot 3$

Multiply $3$ by $h$ to get $3 h$ .

$f (x + h) = x^{2} + x h + 3 x + h x + h^{2} + 3 h + 3 x + 3 h + 3 \cdot 3$

Multiply $3$ by $3$ to get $9$ .

$f (x + h) = x^{2} + x h + 3 x + h x + h^{2} + 3 h + 3 x + 3 h + 9$

Add $x h$ and $h x$ to get $2 h x$ .

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Reorder $x$ and $h$ .

$f (x + h) = x^{2} + (h x + h x) + 3 x + h^{2} + 3 h + 3 x + 3 h + 9$

Add $h x$ and $h x$ to get $2 h x$ .

$f (x + h) = x^{2} + 2 h x + 3 x + h^{2} + 3 h + 3 x + 3 h + 9$

Add $3 x$ and $3 x$ to get $6 x$ .

$f (x + h) = x^{2} + 2 h x + h^{2} + 3 h + 6 x + 3 h + 9$

Add $3 h$ and $3 h$ to get $6 h$ .

$f (x + h) = x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9$

The final answer is $x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9$ .

$x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9$

Find the components of the definition.

$f (x + h) = x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9$

$f (x) = x^{2} + 6 x + 9$

$f (x + h) = x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9$

$f (x) = x^{2} + 6 x + 9$

Plug in the components.

$f' (x) = lim_{h \to 0} \frac{x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9 - (x^{2} + 6 x + 9)}{h}$

Simplify.

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Simplify the numerator.

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Apply the distributive property.

$f' (x) = lim_{h \to 0} \frac{x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9 + (- x^{2} - (6 x) - 1 \cdot 9)}{h}$

Simplify.

$f' (x) = lim_{h \to 0} \frac{x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9 + (- x^{2} - 6 x - 9)}{h}$

Move parentheses.

$f' (x) = lim_{h \to 0} \frac{x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9 + (- x^{2} - 6 x) - 9}{h}$

Remove parentheses.

$f' (x) = lim_{h \to 0} \frac{x^{2} + 2 h x + h^{2} + 6 h + 6 x + 9 - x^{2} - 6 x - 9}{h}$

Subtract $x^{2}$ from $x^{2}$ to get $0$ .

$f' (x) = lim_{h \to 0} \frac{2 h x + h^{2} + 6 h + 6 x + 9 + 0 - 6 x - 9}{h}$

Add $2 h x$ and $0$ to get $2 h x$ .

$f' (x) = lim_{h \to 0} \frac{2 h x + h^{2} + 6 h + 6 x + 9 - 6 x - 9}{h}$

Subtract $6 x$ from $6 x$ to get $0$ .

$f' (x) = lim_{h \to 0} \frac{2 h x + h^{2} + 6 h + 0 + 9 - 9}{h}$

Add $2 h x$ and $0$ to get $2 h x$ .

$f' (x) = lim_{h \to 0} \frac{2 h x + h^{2} + 6 h + 9 - 9}{h}$

Subtract $9$ from $9$ to get $0$ .

$f' (x) = lim_{h \to 0} \frac{2 h x + h^{2} + 6 h + 0}{h}$

Add $2 h x + h^{2} + 6 h$ and $0$ to get $2 h x + h^{2} + 6 h$ .

$f' (x) = lim_{h \to 0} \frac{2 h x + h^{2} + 6 h}{h}$

Factor $h$ out of $2 h x + h^{2} + 6 h$ .

$f' (x) = lim_{h \to 0} \frac{h (2 x + h + 6)}{h}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$f' (x) = lim_{h \to 0} \frac{h (2 x + h + 6)}{h}$

Divide $2 x + h + 6$ by $1$ to get $2 x + h + 6$ .

$f' (x) = lim_{h \to 0} 2 x + h + 6$

Take the limit of each term.

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Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$lim_{h \to 0} 2 x + lim_{h \to 0} h + lim_{h \to 0} 6$

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

$lim_{h \to 0} 2 \cdot lim_{h \to 0} x + lim_{h \to 0} h + lim_{h \to 0} 6$

Evaluate the limits by plugging in $0$ for all occurrences of $h$ .

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Evaluate the limit of $2$ which is constant as $h$ approaches $0$ .

$2 \cdot lim_{h \to 0} x + lim_{h \to 0} h + lim_{h \to 0} 6$

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$2 \cdot x + lim_{h \to 0} h + lim_{h \to 0} 6$

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$2 \cdot x + 0 + lim_{h \to 0} 6$

Evaluate the limit of $6$ which is constant as $h$ approaches $0$ .

$2 \cdot x + 0 + 6$

Simplify the answer.

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Multiply $2$ by $x$ to get $2 x$ .

$2 x + 0 + 6$

Add $2 x$ and $0$ to get $2 x$ .

$2 x + 6$

Find the slope $m$ . In this case $m = 8$ .

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Multiply $2$ by $1$ to get $2$ .

$m = 2 + 6$

Add $2$ and $6$ to get $8$ .

$m = 8$

The slope is $m = 8$ and the point is $(1, 16)$ .

$m = 8, (1, 16)$

Find the value of $b$ using the formula for the equation of a line.

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Use the formula for the equation of a line to find $b$ .

$y = m x + b$

Substitute the value of $m$ into the equation.

$y = (8) \cdot x + b$

Substitute the value of $x$ into the equation.

$y = (8) \cdot (1) + b$

Substitute the value of $y$ into the equation.

$(16) = (8) \cdot (1) + b$

Find the value of $b$ .

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Rewrite the equation as $(8) \cdot (1) + b = 16$ .

$(8) \cdot (1) + b = 16$

Simplify each term.

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Multiply $8$ by $1$ to get $(8) (1)$ .

$(8) (1) + b = 16$

Multiply $8$ by $1$ to get $8$ .

$8 + b = 16$

Move all terms not containing $b$ to the right side of the equation.

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Subtract $8$ from both sides of the equation.

$b = - 8 + 16$

Add $- 8$ and $16$ to get $8$ .

$b = 8$

Now that the values of $m$ (slope) and $b$ (y-intercept) are known, substitute them into $y = m x + b$ to find the equation of the line.

$y = 8 x + 8$

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