Calculus Examples

$y = e^{2 x}$ at $x = \frac{1}{2} \ln (5)$

Step 1

Find the corresponding $y$ -value to $x = \frac{1}{2} \cdot \ln (5)$ .

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Step 1.1

Substitute $\frac{1}{2} \cdot \ln (5)$ in for $x$ .

$y = e^{2 (\frac{1}{2} \cdot \ln (5))}$

Step 1.2

Simplify $e^{2 (\frac{1}{2} \cdot \ln (5))}$ .

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Step 1.2.1

Simplify $\frac{1}{2} \ln (5)$ by moving $\frac{1}{2}$ inside the logarithm.

$y = e^{2 \ln (5^{\frac{1}{2}})}$

Step 1.2.2

Simplify $2 \ln (5^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

$y = e^{\ln ({(5^{\frac{1}{2}})}^{2})}$

Step 1.2.3

Exponentiation and log are inverse functions.

$y = {(5^{\frac{1}{2}})}^{2}$

Step 1.2.4

Multiply the exponents in ${(5^{\frac{1}{2}})}^{2}$ .

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Step 1.2.4.1

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

$y = 5^{\frac{1}{2} \cdot 2}$

Step 1.2.4.2

Cancel the common factor of $2$ .

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Step 1.2.4.2.1

Cancel the common factor.

$y = 5^{\frac{1}{2} \cdot 2}$

Step 1.2.4.2.2

Rewrite the expression.

$y = 5^{1}$

Step 1.2.5

Evaluate the exponent.

$y = 5$

Step 2

Find the first derivative and evaluate at $x = \frac{1}{2} \ln (5)$ and $y = 5$ to find the slope of the tangent line.

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Step 2.1

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

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Step 2.1.1

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

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x]$

Step 2.1.2

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

$e^{u} \frac{d}{d x} [2 x]$

Step 2.1.3

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

$e^{2 x} \frac{d}{d x} [2 x]$

Step 2.2

Differentiate.

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Step 2.2.1

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

$e^{2 x} (2 \frac{d}{d x} [x])$

Step 2.2.2

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

$e^{2 x} (2 \cdot 1)$

Step 2.2.3

Simplify the expression.

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Step 2.2.3.1

Multiply $2$ by $1$ .

$e^{2 x} \cdot 2$

Step 2.2.3.2

Move $2$ to the left of $e^{2 x}$ .

$2 e^{2 x}$

Step 2.3

Evaluate the derivative at $x = \frac{1}{2} \cdot \ln (5)$ .

$2 e^{2 (\frac{1}{2} \cdot \ln (5))}$

Step 2.4

Simplify.

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Step 2.4.1

Simplify $\frac{1}{2} \ln (5)$ by moving $\frac{1}{2}$ inside the logarithm.

$2 e^{2 \ln (5^{\frac{1}{2}})}$

Step 2.4.2

Simplify $2 \ln (5^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

$2 e^{\ln ({(5^{\frac{1}{2}})}^{2})}$

Step 2.4.3

Exponentiation and log are inverse functions.

$2 {(5^{\frac{1}{2}})}^{2}$

Step 2.4.4

Multiply the exponents in ${(5^{\frac{1}{2}})}^{2}$ .

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Step 2.4.4.1

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

$2 \cdot 5^{\frac{1}{2} \cdot 2}$

Step 2.4.4.2

Cancel the common factor of $2$ .

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Step 2.4.4.2.1

Cancel the common factor.

$2 \cdot 5^{\frac{1}{2} \cdot 2}$

Step 2.4.4.2.2

Rewrite the expression.

$2 \cdot 5^{1}$

Step 2.4.5

Evaluate the exponent.

$2 \cdot 5$

Step 2.4.6

Multiply $2$ by $5$ .

$10$

Step 3

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 3.1

Use the slope $10$ and a given point $(\frac{1}{2} \cdot \ln (5), 5)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (5) = 10 \cdot (x - (\frac{1}{2} \cdot \ln (5)))$

Step 3.2

Simplify the equation and keep it in point-slope form.

$y - 5 = 10 \cdot (x - \ln (5^{\frac{1}{2}}))$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $10 \cdot (x - \ln (5^{\frac{1}{2}}))$ .

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Step 3.3.1.1

Rewrite.

$y - 5 = 0 + 0 + 10 \cdot (x - \ln (5^{\frac{1}{2}}))$

Step 3.3.1.2

Simplify by adding zeros.

$y - 5 = 10 \cdot (x - \ln (5^{\frac{1}{2}}))$

Step 3.3.1.3

Apply the distributive property.

$y - 5 = 10 x + 10 (- \ln (5^{\frac{1}{2}}))$

Step 3.3.1.4

Multiply $10 (- \ln (5^{\frac{1}{2}}))$ .

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Step 3.3.1.4.1

Multiply $- 1$ by $10$ .

$y - 5 = 10 x - 10 \ln (5^{\frac{1}{2}})$

Step 3.3.1.4.2

Simplify $- 10 \ln (5^{\frac{1}{2}})$ by moving $10$ inside the logarithm.

$y - 5 = 10 x - \ln ({(5^{\frac{1}{2}})}^{10})$

Step 3.3.1.5

Simplify each term.

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Step 3.3.1.5.1

Multiply the exponents in ${(5^{\frac{1}{2}})}^{10}$ .

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Step 3.3.1.5.1.1

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

$y - 5 = 10 x - \ln (5^{\frac{1}{2} \cdot 10})$

Step 3.3.1.5.1.2

Cancel the common factor of $2$ .

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Step 3.3.1.5.1.2.1

Factor $2$ out of $10$ .

$y - 5 = 10 x - \ln (5^{\frac{1}{2} \cdot (2 (5))})$

Step 3.3.1.5.1.2.2

Cancel the common factor.

$y - 5 = 10 x - \ln (5^{\frac{1}{2} \cdot (2 \cdot 5)})$

Step 3.3.1.5.1.2.3

Rewrite the expression.

$y - 5 = 10 x - \ln (5^{5})$

Step 3.3.1.5.2

Raise $5$ to the power of $5$ .

$y - 5 = 10 x - \ln (3125)$

Step 3.3.2

Add $5$ to both sides of the equation.

$y = 10 x - \ln (3125) + 5$

Step 4