Calculus Examples

$y = e^{2 x}$ at

$x = \frac{1}{2} \ln (2)$

Step 1

Find the corresponding

$y$ -value to

$x = \frac{1}{2} \cdot \ln (2)$ .

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Step 1.1

Substitute

$\frac{1}{2} \cdot \ln (2)$ in for

$x$ .

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

Step 1.2

Simplify

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

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Step 1.2.1

Simplify

$\frac{1}{2} \ln (2)$ by moving

$\frac{1}{2}$ inside the logarithm.

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

Step 1.2.2

Simplify

$2 \ln (2^{\frac{1}{2}})$ by moving

$2$ inside the logarithm.

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

Step 1.2.3

Exponentiation and log are inverse functions.

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

Step 1.2.4

Multiply the exponents in

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

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Step 1.2.4.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

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

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.

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

Step 1.2.4.2.2

Rewrite the expression.

$y = 2^{1}$

Step 1.2.5

Evaluate the exponent.

$y = 2$

Step 2

Find the first derivative and evaluate at

$x = \frac{1}{2} \ln (2)$ and

$y = 2$ 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

$x = \frac{1}{2} \cdot \ln (2)$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Simplify

$\frac{1}{2} \ln (2)$ by moving

$\frac{1}{2}$ inside the logarithm.

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

Step 2.4.2

Simplify

$2 \ln (2^{\frac{1}{2}})$ by moving

$2$ inside the logarithm.

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

Step 2.4.3

Exponentiation and log are inverse functions.

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

Step 2.4.4

Multiply the exponents in

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

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Step 2.4.4.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

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

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.

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

Step 2.4.4.2.2

Rewrite the expression.

$2 \cdot 2^{1}$

Step 2.4.5

Evaluate the exponent.

$2 \cdot 2$

Step 2.4.6

Multiply

$2$ by

$2$ .

$4$

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

$4$ and a given point

$(\frac{1}{2} \cdot \ln (2), 2)$ 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

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (2) = 4 \cdot (x - (\frac{1}{2} \cdot \ln (2)))$

Step 3.2

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

$y - 2 = 4 \cdot (x - \ln (2^{\frac{1}{2}}))$

Step 3.3

Solve for

$y$ .

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Step 3.3.1

Simplify

$4 \cdot (x - \ln (2^{\frac{1}{2}}))$ .

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Step 3.3.1.1

Rewrite.

$y - 2 = 0 + 0 + 4 \cdot (x - \ln (2^{\frac{1}{2}}))$

Step 3.3.1.2

Simplify by adding zeros.

$y - 2 = 4 \cdot (x - \ln (2^{\frac{1}{2}}))$

Step 3.3.1.3

Apply the distributive property.

$y - 2 = 4 x + 4 (- \ln (2^{\frac{1}{2}}))$

Step 3.3.1.4

Multiply

$4 (- \ln (2^{\frac{1}{2}}))$ .

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Step 3.3.1.4.1

Multiply

$- 1$ by

$4$ .

$y - 2 = 4 x - 4 \ln (2^{\frac{1}{2}})$

Step 3.3.1.4.2

Simplify

$- 4 \ln (2^{\frac{1}{2}})$ by moving

$4$ inside the logarithm.

$y - 2 = 4 x - \ln ({(2^{\frac{1}{2}})}^{4})$

Step 3.3.1.5

Simplify each term.

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Step 3.3.1.5.1

Multiply the exponents in

${(2^{\frac{1}{2}})}^{4}$ .

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Step 3.3.1.5.1.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$y - 2 = 4 x - \ln (2^{\frac{1}{2} \cdot 4})$

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

$4$ .

$y - 2 = 4 x - \ln (2^{\frac{1}{2} \cdot (2 (2))})$

Step 3.3.1.5.1.2.2

Cancel the common factor.

$y - 2 = 4 x - \ln (2^{\frac{1}{2} \cdot (2 \cdot 2)})$

Step 3.3.1.5.1.2.3

Rewrite the expression.

$y - 2 = 4 x - \ln (2^{2})$

Step 3.3.1.5.2

Raise

$2$ to the power of

$2$ .

$y - 2 = 4 x - \ln (4)$

Step 3.3.2

Add

$2$ to both sides of the equation.

$y = 4 x - \ln (4) + 2$

Step 4