Question Statement

Find the slope of the tangent to the curve $y = \frac{54}{x ^{2} + 1}$ at the point where $x = 2$ .

Background and Explanation

This problem requires finding the derivative of a rational function to determine the instantaneous rate of change (slope of tangent) at a specific point. You can approach this using either the chain rule after rewriting the function with a negative exponent, or the quotient rule directly.

Solution

To find the slope of the tangent line, we need to compute $\frac{d y}{d x}$ and evaluate it at $x = 2$ .

First, rewrite the function to make differentiation easier using the chain rule:

$y = \frac{54}{x ^{2} + 1} = 54 (x^{2} + 1)^{- 1}$

Now differentiate with respect to $x$ using the chain rule. The chain rule states that if $y = 54 (u)^{- 1}$ where $u = x^{2} + 1$ , then $\frac{d y}{d x} = 54 \cdot (- 1) u^{- 2} \cdot \frac{d u}{d x}$ :

$\frac{d y}{d x} = - 54 (x^{2} + 1)^{- 2} \cdot \frac{d}{d x} (x^{2} + 1)$

Compute the derivative of the inner function $(x^{2} + 1)$ :

$\frac{d y}{d x} = - 54 (x^{2} + 1)^{- 2} \cdot (2 x + 0)$

Simplify the expression:

$\frac{d y}{d x} = \frac{- 54 \cdot 2 x}{( x ^{2} + 1 ) ^{2}} = \frac{- 108 x}{( x ^{2} + 1 ) ^{2}}$

Now evaluate the derivative at $x = 2$ to find the slope at that specific point:

$\frac{d y}{d x}_{x = 2} = \frac{- 108 ( 2 )}{(( 2 ) ^{2} + 1 ) ^{2}} = \frac{- 216}{( 4 + 1 ) ^{2}} = \frac{- 216}{25}$

Therefore, the slope of the tangent to the curve at $x = 2$ is $- \frac{216}{25}$ .

Key Formulas or Methods Used

Chain Rule: $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$
Power Rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$ (extended to negative exponents)
Derivative of a Constant: $\frac{d}{d x} [c] = 0$
Function Rewriting: Converting $\frac{1}{u}$ to $u^{- 1}$ to facilitate chain rule application

Summary of Steps

Rewrite $y = \frac{54}{x ^{2} + 1}$ as $y = 54 (x^{2} + 1)^{- 1}$ to prepare for chain rule differentiation
Apply the chain rule: differentiate the outer function (power $- 1$ ) and multiply by the derivative of the inner function $(x^{2} + 1)$
Simplify to obtain $\frac{d y}{d x} = \frac{- 108 x}{( x ^{2} + 1 ) ^{2}}$
Substitute $x = 2$ into the derivative: $\frac{- 108 ( 2 )}{( 4 + 1 ) ^{2}}$
Calculate the final result: $- \frac{216}{25}$

Solution

To find the slope of the tangent line, we need to compute

\frac{d y}{d x}

and evaluate it at

x = 2

First, rewrite the function to make differentiation easier using the chain rule:

y = \frac{54}{x ^{2} + 1} = 54 (x^{2} + 1)^{- 1}

Now differentiate with respect to

x

using the chain rule. The chain rule states that if

y = 54 (u)^{- 1}

where

u = x^{2} + 1

, then

\frac{d y}{d x} = 54 \cdot (- 1) u^{- 2} \cdot \frac{d u}{d x}

\frac{d y}{d x} = - 54 (x^{2} + 1)^{- 2} \cdot \frac{d}{d x} (x^{2} + 1)

Compute the derivative of the inner function

(x^{2} + 1)

\frac{d y}{d x} = - 54 (x^{2} + 1)^{- 2} \cdot (2 x + 0)

Simplify the expression:

\frac{d y}{d x} = \frac{- 54 \cdot 2 x}{( x ^{2} + 1 ) ^{2}} = \frac{- 108 x}{( x ^{2} + 1 ) ^{2}}

Now evaluate the derivative at

x = 2

to find the slope at that specific point:

\frac{d y}{d x}_{x = 2} = \frac{- 108 ( 2 )}{(( 2 ) ^{2} + 1 ) ^{2}} = \frac{- 216}{( 4 + 1 ) ^{2}} = \frac{- 216}{25}

Therefore, the slope of the tangent to the curve at

x = 2

- \frac{216}{25}

Summary of Steps

Rewrite

y = \frac{54}{x ^{2} + 1}

y = 54 (x^{2} + 1)^{- 1}

to prepare for chain rule differentiation

Apply the chain rule: differentiate the outer function (power

- 1

) and multiply by the derivative of the inner function

(x^{2} + 1)

Simplify to obtain

\frac{d y}{d x} = \frac{- 108 x}{( x ^{2} + 1 ) ^{2}}

Substitute

x = 2

into the derivative:

\frac{- 108 ( 2 )}{( 4 + 1 ) ^{2}}

Calculate the final result:

- \frac{216}{25}

2.5 Q-15

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

2.5 Q-15

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps