tangent of f(x)=(x^2)/(x+1) at x=2

Check the domain before calculating the tangent:

x \neq -1

Find the tangent at

x = 2

⇥1) Let's perform the differentiation:

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

⇥For

p1

x^{2}

and

q2

x + 1

, use the differentiation rule:

\frac{d}{d x} (\frac{p1}{q2}) = \frac{\frac{d}{d x} (p1) \cdot q2 - p1 \cdot \frac{d}{d x} (q2)}{{q2}^{2}}

⇥2) Let's differentiate

p1

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

⇥ⓘ The derivative of x^n is n*x^(n-1):

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

2 x

⇥

2 x

⇥3) Let's perform the differentiation

q2

\frac{d}{d x} (x + 1)

⇥ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x + 1)

1 + 0

⇥

1 + 0

⇥ⓘ Adding 0 changes nothing:

1 + 0

1

⇥

1

⇥4) Let's evaluate the resulting derivative (

\frac{p1}{q2}

)':

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

⇥

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

⇥ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

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

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

⇥

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

⇥ⓘ Expand by multiplying each term inside the parentheses:

2 \cdot (x + 1)

2 x + 2

⇥

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

⇥ⓘ Like terms are grouped by adding their coefficients:

2 x - x

x

⇥

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

⇥5) Then you can continue the derivation:

⇥ⅰ

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

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

⇥

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

Compute f(

2

)

⇥

\frac{2^{2}}{2 + 1}

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

■The numerator and the denominator are computed before the division:

2 + 1

3

⇥

\frac{4}{3}

The point of tangency is (

2

\frac{4}{3}

)

Compute f'(

2

)

⇥

\frac{2 \cdot (2 + 2)}{{(2 + 1)}^{2}}

⇥ⓘ Parentheses are computed first:

2 + 2

4

2 + 1

3

⇥

\frac{2 \cdot 4}{3^{2}}

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

3^{2}

3 \cdot 3

9

⇥ⅰ

2 \cdot 4

8

⇥

\frac{8}{9}

The slope is f'(

2

) =

\frac{8}{9}

The tangent passes through this point with this slope:

y = \frac{8}{9} \cdot (x - 2) + \frac{4}{3}

Compute the y-intercept:

b = \frac{4}{3} - \frac{8}{9} \cdot 2

⇥

\frac{4}{3} - \frac{8}{9} \cdot 2

⇥ⓘ Multiplying a fraction by a number multiplies its numerator:

\frac{8}{9} \cdot 2

\frac{8 \cdot 2}{9}

\frac{16}{9}

⇥

\frac{4}{3} - \frac{16}{9}

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators. We choose

9

, the least common multiple of

3

and

9

. Multiplying the numerator and the denominator by the same number does not change the fraction:

\frac{4}{3} - \frac{16}{9}

\frac{4 \cdot 3}{3 \cdot 3} - \frac{16}{9}

\frac{4 \cdot 3 - 16}{9}

⇥

\frac{4 \cdot 3 - 16}{9}

⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 3

12

⇥

\frac{12 - 16}{9}

⇥ⓘ The numerator and the denominator are computed before the division:

12 - 16

-4

⇥

- \frac{4}{9}

⇥

y = \frac{8}{9} \cdot x - \frac{4}{9}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{8}{9} \cdot x

\frac{8 x}{9}

■To add fractions, put them over the same denominator, then add the numerators:

\frac{8 x}{9} - \frac{4}{9}

\frac{8 x - 4}{9}

▶Tangent line:

y = \frac{8 x - 4}{9}