Find the Tangent Line at x=6 f(x) = square root of x+3 , x=6

$f (x) = \sqrt{x + 3}$ , $x = 6$

Step 1

Find the corresponding $y$ -value to $x = 6$ .

Substitute $6$ in for $x$ .

$y = \sqrt{(6) + 3}$

Simplify $\sqrt{(6) + 3}$ .

Add $6$ and $3$ .

$y = \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$y = \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$y = 3$

Step 2

Find the first derivative and evaluate at $x = 6$ and $y = 3$ to find the slope of the tangent line.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 3}$ as ${(x + 3)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [{(x + 3)}^{\frac{1}{2}}]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x + 3$ .

To apply the Chain Rule, set $u$ as $x + 3$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x + 3]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x + 3]$

Replace all occurrences of $u$ with $x + 3$ .

$\frac{1}{2} {(x + 3)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 3]$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} {(x + 3)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x + 3]$

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{1}{2} {(x + 3)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x + 3]$

Combine the numerators over the common denominator.

$\frac{1}{2} {(x + 3)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x + 3]$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(x + 3)}^{\frac{1 - 2}{2}} \frac{d}{d x} [x + 3]$

Subtract $2$ from $1$ .

$\frac{1}{2} {(x + 3)}^{\frac{- 1}{2}} \frac{d}{d x} [x + 3]$

Combine fractions.

Move the negative in front of the fraction.

$\frac{1}{2} {(x + 3)}^{- \frac{1}{2}} \frac{d}{d x} [x + 3]$

Combine $\frac{1}{2}$ and ${(x + 3)}^{- \frac{1}{2}}$ .

$\frac{{(x + 3)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [x + 3]$

Move ${(x + 3)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(x + 3)}^{\frac{1}{2}}} \frac{d}{d x} [x + 3]$

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$\frac{1}{2 {(x + 3)}^{\frac{1}{2}}} (\frac{d}{d x} [x] + \frac{d}{d x} [3])$

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

$\frac{1}{2 {(x + 3)}^{\frac{1}{2}}} (1 + \frac{d}{d x} [3])$

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$\frac{1}{2 {(x + 3)}^{\frac{1}{2}}} (1 + 0)$

Simplify the expression.

Add $1$ and $0$ .

$\frac{1}{2 {(x + 3)}^{\frac{1}{2}}} \cdot 1$

Multiply $\frac{1}{2 {(x + 3)}^{\frac{1}{2}}}$ by $1$ .

$\frac{1}{2 {(x + 3)}^{\frac{1}{2}}}$

Evaluate the derivative at $x = 6$ .

$\frac{1}{2 {((6) + 3)}^{\frac{1}{2}}}$

Simplify.

Simplify the denominator.

Add $6$ and $3$ .

$\frac{1}{2 \cdot 9^{\frac{1}{2}}}$

Rewrite $9$ as $3^{2}$ .

$\frac{1}{2 \cdot {(3^{2})}^{\frac{1}{2}}}$

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

$\frac{1}{2 \cdot 3^{2 (\frac{1}{2})}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{1}{2 \cdot 3^{2 (\frac{1}{2})}}$

Rewrite the expression.

$\frac{1}{2 \cdot 3^{1}}$

Evaluate the exponent.

$\frac{1}{2 \cdot 3}$

Multiply $2$ by $3$ .

$\frac{1}{6}$

Step 3

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

Use the slope $\frac{1}{6}$ and a given point $(6, 3)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (3) = \frac{1}{6} \cdot (x - (6))$

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

$y - 3 = \frac{1}{6} \cdot (x - 6)$

Solve for $y$ .

Simplify $\frac{1}{6} \cdot (x - 6)$ .

Rewrite.

$y - 3 = 0 + 0 + \frac{1}{6} \cdot (x - 6)$

Simplify by adding zeros.

$y - 3 = \frac{1}{6} \cdot (x - 6)$

Apply the distributive property.

$y - 3 = \frac{1}{6} x + \frac{1}{6} \cdot - 6$

Combine $\frac{1}{6}$ and $x$ .

$y - 3 = \frac{x}{6} + \frac{1}{6} \cdot - 6$

Cancel the common factor of $6$ .

Factor $6$ out of $- 6$ .

$y - 3 = \frac{x}{6} + \frac{1}{6} \cdot (6 (- 1))$

Cancel the common factor.

$y - 3 = \frac{x}{6} + \frac{1}{6} \cdot (6 \cdot - 1)$

Rewrite the expression.

$y - 3 = \frac{x}{6} - 1$

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

Add $3$ to both sides of the equation.

$y = \frac{x}{6} - 1 + 3$

Add $- 1$ and $3$ .

$y = \frac{x}{6} + 2$

Reorder terms.

$y = \frac{1}{6} x + 2$

Step 4

Find the Tangent Line at x=6 f(x) = square root of x+3 , x=6

Related Questions

Support LCMGCF.com by disabling your adblocker.