Calculus Examples

y = ln (x^{2})

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

(2, ln (4))

$(2, \ln (4))$

Step 1

Find the first derivative and evaluate at

x = 2

$x = 2$ and

y = ln (4)

$y = \ln (4)$ to find the slope of the tangent line.

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Step 1.1

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = ln (x)

$f (x) = \ln (x)$ and

g (x) = x^{2}

$g (x) = x^{2}$ .

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Step 1.1.1

To apply the Chain Rule, set

u

$u$ as

x^{2}

$x^{2}$ .

\frac{d}{d u} [ln (u)] \frac{d}{d x} [x^{2}]

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

Step 1.1.2

The derivative of

ln (u)

$\ln (u)$ with respect to

u

$u$ is

\frac{1}{u}

$\frac{1}{u}$ .

\frac{1}{u} \frac{d}{d x} [x^{2}]

$\frac{1}{u} \frac{d}{d x} [x^{2}]$

Step 1.1.3

Replace all occurrences of

u

$u$ with

x^{2}

$x^{2}$ .

\frac{1}{x^{2}} \frac{d}{d x} [x^{2}]

$\frac{1}{x^{2}} \frac{d}{d x} [x^{2}]$

\frac{1}{x^{2}} \frac{d}{d x} [x^{2}]

$\frac{1}{x^{2}} \frac{d}{d x} [x^{2}]$

Step 1.2

Differentiate using the Power Rule.

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Step 1.2.1

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 2

$n = 2$ .

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

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

Step 1.2.2

Simplify terms.

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Step 1.2.2.1

Combine

$2$ and

$\frac{1}{x^{2}}$ .

$\frac{2}{x^{2}} x$

Step 1.2.2.2

Combine

$\frac{2}{x^{2}}$ and

$x$ .

$\frac{2 x}{x^{2}}$

Step 1.2.2.3

Cancel the common factor of

$x$ and

$x^{2}$ .

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Step 1.2.2.3.1

Factor

$x$ out of

$2 x$ .

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

Step 1.2.2.3.2

Cancel the common factors.

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Step 1.2.2.3.2.1

Factor

$x$ out of

$x^{2}$ .

$\frac{x \cdot 2}{x \cdot x}$

Step 1.2.2.3.2.2

Cancel the common factor.

$\frac{x \cdot 2}{x \cdot x}$

Step 1.2.2.3.2.3

Rewrite the expression.

$\frac{2}{x}$

Step 1.3

Evaluate the derivative at

$x = 2$ .

$\frac{2}{2}$

Step 1.4

Divide

$2$ by

$2$ .

$1$

Step 2

Plug the slope and point values into the point-slope formula and solve for

$y$ .

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Step 2.1

Use the slope

$1$ and a given point

$(2, \ln (4))$ 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

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

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

Step 2.2

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

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

Step 2.3

Solve for

$y$ .

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Step 2.3.1

Multiply

$x - 2$ by

$1$ .

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

Step 2.3.2

Add

$\ln (4)$ to both sides of the equation.

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

Step 3