Find the Tangent Line at the Point 3y^2- square root of x=44 , (16,4)

$3 y^{2} - \sqrt{x} = 44$ , $(16, 4)$

Step 1

Find the first derivative and evaluate at $x = 16$ and $y = 4$ to find the slope of the tangent line.

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

$3 y^{2} - x^{\frac{1}{2}} = 44$

Differentiate both sides of the equation.

$\frac{d}{d x} (3 y^{2} - x^{\frac{1}{2}}) = \frac{d}{d x} (44)$

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $3 y^{2} - x^{\frac{1}{2}}$ with respect to $x$ is $\frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [- x^{\frac{1}{2}}]$ .

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

Evaluate $\frac{d}{d x} [3 y^{2}]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 y^{2}$ with respect to $x$ is $3 \frac{d}{d x} [y^{2}]$ .

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

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

To apply the Chain Rule, set $u$ as $y$ .

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

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

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

Replace all occurrences of $u$ with $y$ .

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

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

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

Multiply $2$ by $3$ .

$6 y y' + \frac{d}{d x} [- x^{\frac{1}{2}}]$

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{\frac{1}{2}}$ with respect to $x$ is $- \frac{d}{d x} [x^{\frac{1}{2}}]$ .

$6 y y' - \frac{d}{d x} [x^{\frac{1}{2}}]$

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

$6 y y' - (\frac{1}{2} x^{\frac{1}{2} - 1})$

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

$6 y y' - (\frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}})$

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

$6 y y' - (\frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}})$

Combine the numerators over the common denominator.

$6 y y' - (\frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}})$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$6 y y' - (\frac{1}{2} x^{\frac{1 - 2}{2}})$

Subtract $2$ from $1$ .

$6 y y' - (\frac{1}{2} x^{\frac{- 1}{2}})$

Move the negative in front of the fraction.

$6 y y' - (\frac{1}{2} x^{- \frac{1}{2}})$

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

$6 y y' - \frac{x^{- \frac{1}{2}}}{2}$

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

$6 y y' - \frac{1}{2 x^{\frac{1}{2}}}$

Since $44$ is constant with respect to $x$ , the derivative of $44$ with respect to $x$ is $0$ .

$0$

Reform the equation by setting the left side equal to the right side.

$6 y y' - \frac{1}{2 x^{\frac{1}{2}}} = 0$

Solve for $y'$ .

Add $\frac{1}{2 x^{\frac{1}{2}}}$ to both sides of the equation.

$6 y y' = \frac{1}{2 x^{\frac{1}{2}}}$

Divide each term in $6 y y' = \frac{1}{2 x^{\frac{1}{2}}}$ by $6 y$ and simplify.

Divide each term in $6 y y' = \frac{1}{2 x^{\frac{1}{2}}}$ by $6 y$ .

$\frac{6 y y'}{6 y} = \frac{\frac{1}{2 x^{\frac{1}{2}}}}{6 y}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 y y'}{6 y} = \frac{\frac{1}{2 x^{\frac{1}{2}}}}{6 y}$

Rewrite the expression.

$\frac{y y'}{y} = \frac{\frac{1}{2 x^{\frac{1}{2}}}}{6 y}$

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{y y'}{y} = \frac{\frac{1}{2 x^{\frac{1}{2}}}}{6 y}$

Divide $y'$ by $1$ .

$y' = \frac{\frac{1}{2 x^{\frac{1}{2}}}}{6 y}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$y' = \frac{1}{2 x^{\frac{1}{2}}} \cdot \frac{1}{6 y}$

Combine.

$y' = \frac{1 \cdot 1}{2 x^{\frac{1}{2}} (6 y)}$

Simplify the expression.

Multiply $1$ by $1$ .

$y' = \frac{1}{2 x^{\frac{1}{2}} \cdot 6 y}$

Multiply $6$ by $2$ .

$y' = \frac{1}{12 x^{\frac{1}{2}} y}$

Reorder factors in $\frac{1}{12 x^{\frac{1}{2}} y}$ .

$y' = \frac{1}{12 y x^{\frac{1}{2}}}$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{1}{12 y x^{\frac{1}{2}}}$

Evaluate at $x = 16$ and $y = 4$ .

Replace the variable $x$ with $16$ in the expression.

$\frac{1}{12 y {(16)}^{\frac{1}{2}}}$

Replace the variable $y$ with $4$ in the expression.

$\frac{1}{12 (4) {(16)}^{\frac{1}{2}}}$

Multiply $12$ by $4$ .

$\frac{1}{48 \cdot 16^{\frac{1}{2}}}$

Simplify the denominator.

Rewrite $16$ as $4^{2}$ .

$\frac{1}{48 \cdot {(4^{2})}^{\frac{1}{2}}}$

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

$\frac{1}{48 \cdot 4^{2 (\frac{1}{2})}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{1}{48 \cdot 4^{2 (\frac{1}{2})}}$

Rewrite the expression.

$\frac{1}{48 \cdot 4^{1}}$

Evaluate the exponent.

$\frac{1}{48 \cdot 4}$

Multiply $48$ by $4$ .

$\frac{1}{192}$

Step 2

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

Use the slope $\frac{1}{192}$ and a given point $(16, 4)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (4) = \frac{1}{192} \cdot (x - (16))$

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

$y - 4 = \frac{1}{192} \cdot (x - 16)$

Solve for $y$ .

Simplify $\frac{1}{192} \cdot (x - 16)$ .

Rewrite.

$y - 4 = 0 + 0 + \frac{1}{192} \cdot (x - 16)$

Simplify by adding zeros.

$y - 4 = \frac{1}{192} \cdot (x - 16)$

Apply the distributive property.

$y - 4 = \frac{1}{192} x + \frac{1}{192} \cdot - 16$

Combine $\frac{1}{192}$ and $x$ .

$y - 4 = \frac{x}{192} + \frac{1}{192} \cdot - 16$

Cancel the common factor of $16$ .

Factor $16$ out of $192$ .

$y - 4 = \frac{x}{192} + \frac{1}{16 (12)} \cdot - 16$

Factor $16$ out of $- 16$ .

$y - 4 = \frac{x}{192} + \frac{1}{16 \cdot 12} \cdot (16 \cdot - 1)$

Cancel the common factor.

$y - 4 = \frac{x}{192} + \frac{1}{16 \cdot 12} \cdot (16 \cdot - 1)$

Rewrite the expression.

$y - 4 = \frac{x}{192} + \frac{1}{12} \cdot - 1$

Combine $\frac{1}{12}$ and $- 1$ .

$y - 4 = \frac{x}{192} + \frac{- 1}{12}$

Move the negative in front of the fraction.

$y - 4 = \frac{x}{192} - \frac{1}{12}$

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

Add $4$ to both sides of the equation.

$y = \frac{x}{192} - \frac{1}{12} + 4$

To write $4$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$y = \frac{x}{192} - \frac{1}{12} + 4 \cdot \frac{12}{12}$

Combine $4$ and $\frac{12}{12}$ .

$y = \frac{x}{192} - \frac{1}{12} + \frac{4 \cdot 12}{12}$

Combine the numerators over the common denominator.

$y = \frac{x}{192} + \frac{- 1 + 4 \cdot 12}{12}$

Simplify the numerator.

Multiply $4$ by $12$ .

$y = \frac{x}{192} + \frac{- 1 + 48}{12}$

Add $- 1$ and $48$ .

$y = \frac{x}{192} + \frac{47}{12}$

Reorder terms.

$y = \frac{1}{192} x + \frac{47}{12}$

Step 3

Find the Tangent Line at the Point 3y^2- square root of x=44 , (16,4)

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