tangent of y=(-8x)/(x^2+1),(-1,4)

⚠ 8x is read as the product 8*x: for a power, write 8^x

1) Check the domain before calculating the tangent:

x \in ℝ

Find the tangent at

x = -1

⇥1.1) Let's perform the differentiation:

\frac{d}{d x} (- \frac{8 x}{x^{2} + 1})

⇥We differentiate

u1

8 x

and

v2

x^{2} + 1

using:

\frac{d}{d x} (\frac{u1}{v2}) = \frac{\frac{d}{d x} (u1) \cdot v2 - u1 \cdot \frac{d}{d x} (v2)}{{v2}^{2}}

⇥1.2) Let's perform the differentiation

u1

\frac{d}{d x} (8 x)

⇥

8

⇥1.3) Let's differentiate

v2

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

⇥

\frac{d}{d x} (x^{2}) + 0

⇥

2 x

⇥1.4) You can evaluate the resulted derivative (

\frac{u1}{v2}

)':

- \frac{8 \cdot (x^{2} + 1) - 8 x \cdot 2 x}{{(x^{2} + 1)}^{2}}

⇥ℹ

8 \cdot 2

16

‖

x \cdot x

x^{2}

⇥

- \frac{8 \cdot (x^{2} + 1) - 16 x^{2}}{{(x^{2} + 1)}^{2}}

⇥

\frac{- 8 \cdot (x^{2} + 1) + 16 x^{2}}{{(x^{2} + 1)}^{2}}

⇥

\frac{- (8 x^{2} + 8) + 16 x^{2}}{{(x^{2} + 1)}^{2}}

⇥

\frac{- 8 x^{2} - 8 + 16 x^{2}}{{(x^{2} + 1)}^{2}}

⇥

\frac{8 x^{2} - 8}{{(x^{2} + 1)}^{2}}

Compute f(

-1

)

⇥

- \frac{8 \cdot (-1)}{{(-1)}^{2} + 1}

⇥ℹ

8 \cdot (-1)

-8

‖

{(-1)}^{2}

1

⇥

- \frac{-8}{1 + 1}

⇥

\frac{8}{1 + 1}

⇥

\frac{8}{2}

⇥

4

The point of tangency is (

-1

4

)

Compute f'(

-1

)

⇥

\frac{8 \cdot {(-1)}^{2} - 8}{{({(-1)}^{2} + 1)}^{2}}

⇥ℹ

{(-1)}^{2}

1

‖

8 \cdot 1

8

⇥

\frac{8 - 8}{{(1 + 1)}^{2}}

⇥ℹ

8 - 8

0

‖

1 + 1

2

⇥

\frac{0}{2^{2}}

⇥

0

The slope is f'(

-1

) =

0

The tangent passes through this point with this slope:

y = 0 \cdot (x + 1) + 4

Compute the y-intercept:

b = 4 - 0 \cdot (-1)

⇥

4 - 0 \cdot (-1)

⇥

4 + 0

⇥

4

⇥

y = 0 \cdot x + 4

⇥

y = 0 + 4

▶Tangent line:

y = 4