Question Statement

Find the slope of the tangent to the curve $y = \frac{2 x + 5}{x + 2}$ at $x = 1$ .

Background and Explanation

This problem requires differentiating a rational function using the quotient rule. The slope of the tangent line to a curve at a specific point equals the value of the derivative evaluated at that point.

Solution

Given the function:

$y = \frac{2 x + 5}{x + 2}$

Step 1: Differentiate using the Quotient Rule

Differentiating with respect to $x$ :

$\frac{d y}{d x} = \frac{( x + 2 ) \frac{d}{d x} ( 2 x + 5 ) - ( 2 x + 5 ) \frac{d}{d x} ( x + 2 )}{( x + 2 ) ^{2}}$

Step 2: Compute the individual derivatives

Substituting the derivatives of the numerator and denominator:

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

Step 3: Simplify the expression

Expanding the terms in the numerator:

$= \frac{2 x + 4 - 2 x - 5}{( x + 2 ) ^{2}}$

Combining like terms ( $2 x - 2 x = 0$ and $4 - 5 = - 1$ ):

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

Therefore, the general derivative is:

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

Step 4: Evaluate at $x = 1$

Substituting $x = 1$ into the derivative:

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

$= \frac{- 1}{( 3 ) ^{2}}$

$= \frac{- 1}{9}$

Thus, the slope of the tangent at $x = 1$ is:

$\frac{d y}{d x} = \frac{- 1}{9}$

Key Formulas or Methods Used

Quotient Rule: $\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ where $u$ and $v$ are differentiable functions of $x$
Linear Differentiation: $\frac{d}{d x} (a x + b) = a$
Geometric Interpretation: The derivative $\frac{d y}{d x}$ at a point represents the slope of the tangent line to the curve at that point

Summary of Steps

Identify the numerator $u = 2 x + 5$ and denominator $v = x + 2$
Apply the quotient rule formula: $\frac{d y}{d x} = \frac{v \cdot u ^{'} - u \cdot v ^{'}}{v ^{2}}$
Differentiate: $u^{'} = 2$ and $v^{'} = 1$
Substitute and expand: $\frac{( x + 2 ) ( 2 ) - ( 2 x + 5 ) ( 1 )}{( x + 2 ) ^{2}} = \frac{2 x + 4 - 2 x - 5}{( x + 2 ) ^{2}}$
Simplify to get: $\frac{d y}{d x} = \frac{- 1}{( x + 2 ) ^{2}}$
Evaluate at $x = 1$ : $\frac{- 1}{( 3 ) ^{2}} = - \frac{1}{9}$
Conclude that the slope of the tangent line is $- \frac{1}{9}$

Question Statement

Find the slope of the tangent to the curve $y = \frac{2 x + 5}{x + 2}$ at $x = 1$ .

Background and Explanation

Solution

Given the function:

$y = \frac{2 x + 5}{x + 2}$

Step 1: Differentiate using the Quotient Rule

Differentiating with respect to $x$ :

$\frac{d y}{d x} = \frac{( x + 2 ) \frac{d}{d x} ( 2 x + 5 ) - ( 2 x + 5 ) \frac{d}{d x} ( x + 2 )}{( x + 2 ) ^{2}}$

Step 2: Compute the individual derivatives

Substituting the derivatives of the numerator and denominator:

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

Step 3: Simplify the expression

Expanding the terms in the numerator:

$= \frac{2 x + 4 - 2 x - 5}{( x + 2 ) ^{2}}$

Combining like terms ( $2 x - 2 x = 0$ and $4 - 5 = - 1$ ):

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

Therefore, the general derivative is:

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

Step 4: Evaluate at $x = 1$

Substituting $x = 1$ into the derivative:

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

$= \frac{- 1}{( 3 ) ^{2}}$

$= \frac{- 1}{9}$

Thus, the slope of the tangent at $x = 1$ is:

$\frac{d y}{d x} = \frac{- 1}{9}$

Key Formulas or Methods Used

Quotient Rule: $\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ where $u$ and $v$ are differentiable functions of $x$
Linear Differentiation: $\frac{d}{d x} (a x + b) = a$
Geometric Interpretation: The derivative $\frac{d y}{d x}$ at a point represents the slope of the tangent line to the curve at that point

Summary of Steps

Identify the numerator $u = 2 x + 5$ and denominator $v = x + 2$
Apply the quotient rule formula: $\frac{d y}{d x} = \frac{v \cdot u ^{'} - u \cdot v ^{'}}{v ^{2}}$
Differentiate: $u^{'} = 2$ and $v^{'} = 1$
Substitute and expand: $\frac{( x + 2 ) ( 2 ) - ( 2 x + 5 ) ( 1 )}{( x + 2 ) ^{2}} = \frac{2 x + 4 - 2 x - 5}{( x + 2 ) ^{2}}$
Simplify to get: $\frac{d y}{d x} = \frac{- 1}{( x + 2 ) ^{2}}$
Evaluate at $x = 1$ : $\frac{- 1}{( 3 ) ^{2}} = - \frac{1}{9}$
Conclude that the slope of the tangent line is $- \frac{1}{9}$

2.5 Q-16

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

2.5 Q-16

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps