Calculus Examples

$7 x^{2} + 3 x$ , $(1, 10)$

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

$m$ $=$ The derivative of $7 x^{2} + 3 x$

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.

$f (x + h) = 7 {(x + h)}^{2} + 3 (x + h)$

Simplify the result.

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Simplify each term.

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

$f (x + h) = 7 ((x + h) (x + h)) + 3 (x + h)$

Expand $(x + h) (x + h)$ using the FOIL Method.

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Apply the distributive property.

$f (x + h) = 7 (x (x + h) + h (x + h)) + 3 (x + h)$

Apply the distributive property.

$f (x + h) = 7 (x \cdot x + x h + h (x + h)) + 3 (x + h)$

Apply the distributive property.

$f (x + h) = 7 (x \cdot x + x h + h x + h \cdot h) + 3 (x + h)$

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ .

$f (x + h) = 7 (x^{2} + x h + h x + h \cdot h) + 3 (x + h)$

Multiply $h$ by $h$ .

$f (x + h) = 7 (x^{2} + x h + h x + h^{2}) + 3 (x + h)$

Add $x h$ and $h x$ .

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Reorder $x$ and $h$ .

$f (x + h) = 7 (x^{2} + h x + h x + h^{2}) + 3 (x + h)$

Add $h x$ and $h x$ .

$f (x + h) = 7 (x^{2} + 2 h x + h^{2}) + 3 (x + h)$

Apply the distributive property.

$f (x + h) = 7 x^{2} + 7 (2 h x) + 7 h^{2} + 3 (x + h)$

Multiply $2$ by $7$ .

$f (x + h) = 7 x^{2} + 14 (h x) + 7 h^{2} + 3 (x + h)$

Apply the distributive property.

$f (x + h) = 7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h$

The final answer is $7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h$ .

$7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h$

Find the components of the definition.

$f (x + h) = 7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h$

$f (x) = 7 x^{2} + 3 x$

$f (x + h) = 7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h$

$f (x) = 7 x^{2} + 3 x$

Plug in the components.

$f' (x) = lim_{h \to 0} \frac{7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h - (7 x^{2} + 3 x)}{h}$

Simplify.

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Simplify the numerator.

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Apply the distributive property.

$f' (x) = lim_{h \to 0} \frac{7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h - (7 x^{2}) - (3 x)}{h}$

Multiply $7$ by $- 1$ .

$f' (x) = lim_{h \to 0} \frac{7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h - 7 x^{2} - (3 x)}{h}$

Multiply $3$ by $- 1$ .

$f' (x) = lim_{h \to 0} \frac{7 x^{2} + 14 h x + 7 h^{2} + 3 x + 3 h - 7 x^{2} - 3 x}{h}$

Subtract $7 x^{2}$ from $7 x^{2}$ .

$f' (x) = lim_{h \to 0} \frac{14 h x + 7 h^{2} + 3 x + 3 h + 0 - 3 x}{h}$

Add $14 h x$ and $0$ .

$f' (x) = lim_{h \to 0} \frac{14 h x + 7 h^{2} + 3 x + 3 h - 3 x}{h}$

Subtract $3 x$ from $3 x$ .

$f' (x) = lim_{h \to 0} \frac{14 h x + 7 h^{2} + 3 h + 0}{h}$

Add $14 h x + 7 h^{2} + 3 h$ and $0$ .

$f' (x) = lim_{h \to 0} \frac{14 h x + 7 h^{2} + 3 h}{h}$

Factor $h$ out of $14 h x + 7 h^{2} + 3 h$ .

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Factor $h$ out of $14 h x$ .

$f' (x) = lim_{h \to 0} \frac{h (14 x) + 7 h^{2} + 3 h}{h}$

Factor $h$ out of $7 h^{2}$ .

$f' (x) = lim_{h \to 0} \frac{h (14 x) + h (7 h) + 3 h}{h}$

Factor $h$ out of $3 h$ .

$f' (x) = lim_{h \to 0} \frac{h (14 x) + h (7 h) + h \cdot 3}{h}$

Factor $h$ out of $h (14 x) + h (7 h)$ .

$f' (x) = lim_{h \to 0} \frac{h (14 x + 7 h) + h \cdot 3}{h}$

Factor $h$ out of $h (14 x + 7 h) + h \cdot 3$ .

$f' (x) = lim_{h \to 0} \frac{h (14 x + 7 h + 3)}{h}$

Divide $14 x + 7 h + 3$ by $1$ .

$f' (x) = lim_{h \to 0} 14 x + 7 h + 3$

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 $0$ .

$lim_{h \to 0} 14 x + lim_{h \to 0} 7 h + lim_{h \to 0} 3$

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

$lim_{h \to 0} 14 \cdot lim_{h \to 0} x + lim_{h \to 0} 7 h + lim_{h \to 0} 3$

Move the term $7$ outside of the limit because it is constant with respect to $h$ .

$lim_{h \to 0} 14 \cdot lim_{h \to 0} x + 7 lim_{h \to 0} h + lim_{h \to 0} 3$

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

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Evaluate the limit of $14$ which is constant as $h$ approaches $0$ .

$14 \cdot lim_{h \to 0} x + 7 lim_{h \to 0} h + lim_{h \to 0} 3$

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

$14 \cdot x + 7 lim_{h \to 0} h + lim_{h \to 0} 3$

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

$14 \cdot x + 7 \cdot 0 + lim_{h \to 0} 3$

Evaluate the limit of $3$ which is constant as $h$ approaches $0$ .

$14 \cdot x + 7 \cdot 0 + 3$

Simplify the answer.

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Multiply $7$ by $0$ .

$14 x + 0 + 3$

Add $14 x$ and $0$ .

$14 x + 3$

Simplify $14 (1) + 3$ .

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Multiply $14$ by $1$ .

$m = 14 + 3$

Add $14$ and $3$ .

$m = 17$

The slope is $m = 17$ and the point is $(1, 10)$ .

$m = 17, (1, 10)$

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.

$y = (17) \cdot x + b$

Substitute the value of $x$ into the equation.

$y = (17) \cdot (1) + b$

Substitute the value of $y$ into the equation.

$10 = (17) \cdot (1) + b$

Find the value of $b$ .

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Rewrite the equation as $(17) \cdot (1) + b = 10$ .

$(17) \cdot (1) + b = 10$

Multiply $17$ by $1$ .

$17 + b = 10$

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

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Subtract $17$ from both sides of the equation.

$b = 10 - 17$

Subtract $17$ from $10$ .

$b = - 7$

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.

$y = 17 x - 7$

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