derivative of ((x^2+1^2)/((x^4+1)^4))

1) Let's differentiate:

\frac{d}{d x} (\frac{x^{2} + 1^{2}}{{(x^{4} + 1)}^{4}})

For

p1

x^{2} + 1

and

q2

{(x^{4} + 1)}^{4}

, 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 perform the differentiation

p1

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

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

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

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

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

ⓘ Adding 0 changes nothing:

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

\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^{4} + 1)}^{4})

Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

with f =

x^{4} + 1

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d x} ({(x^{4} + 1)}^{4})

4 \cdot \frac{d}{d x} (x^{4} + 1) \cdot {(x^{4} + 1)}^{3}

4 \cdot \frac{d}{d x} (x^{4} + 1) \cdot {(x^{4} + 1)}^{3}

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

\frac{d}{d x} (x^{4} + 1)

\frac{d}{d x} (x^{4}) + 0

4 \cdot (\frac{d}{d x} (x^{4}) + 0) \cdot {(x^{4} + 1)}^{3}

ⓘ Adding 0 changes nothing:

\frac{d}{d x} (x^{4}) + 0

\frac{d}{d x} (x^{4})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{4})

4 x^{3}

4 \cdot 4 x^{3} \cdot {(x^{4} + 1)}^{3}

ⅰ

4 \cdot 4

16

16 x^{3} \cdot {(x^{4} + 1)}^{3}

4) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{2 x \cdot {(x^{4} + 1)}^{4} - (x^{2} + 1) \cdot 16 x^{3} \cdot {(x^{4} + 1)}^{3}}{{({(x^{4} + 1)}^{4})}^{2}}

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

\frac{2 x \cdot {(x^{4} + 1)}^{4} - (x^{2} + 1) \cdot 16 x^{3} \cdot {(x^{4} + 1)}^{3}}{{({(x^{4} + 1)}^{4})}^{2}}

\frac{2 x \cdot ({(x^{4} + 1)}^{4} - 8 x^{2} \cdot (x^{2} + 1) \cdot {(x^{4} + 1)}^{3})}{{({(x^{4} + 1)}^{4})}^{2}}

■A power of a power multiplies the exponents, (a^n)^m = a^(n*m):

{({(x^{4} + 1)}^{4})}^{2}

{(x^{4} + 1)}^{4 \cdot 2}

\frac{2 x \cdot ({(x^{4} + 1)}^{4} - 8 x^{2} \cdot (x^{2} + 1) \cdot {(x^{4} + 1)}^{3})}{{(x^{4} + 1)}^{4 \cdot 2}}

ⅰ

4 \cdot 2

8

\frac{2 x \cdot ({(x^{4} + 1)}^{4} - 8 x^{2} \cdot (x^{2} + 1) \cdot {(x^{4} + 1)}^{3})}{{(x^{4} + 1)}^{8}}

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

\frac{2 x \cdot ({(x^{4} + 1)}^{4} - 8 x^{2} \cdot (x^{2} + 1) \cdot {(x^{4} + 1)}^{3})}{{(x^{4} + 1)}^{8}}

\frac{2 x \cdot {(x^{4} + 1)}^{3} \cdot ((x^{4} + 1) - 8 x^{2} \cdot (x^{2} + 1))}{{(x^{4} + 1)}^{8}}

\frac{2 x \cdot {(x^{4} + 1)}^{3} \cdot ((x^{4} + 1) - 8 x^{2} \cdot (x^{2} + 1))}{{(x^{4} + 1)}^{8}}

The fraction:

\frac{{(x^{4} + 1)}^{3}}{{(x^{4} + 1)}^{8}}

simplifies by the common factor:

{(x^{4} + 1)}^{3}

into the new quotient:

\frac{1}{{(x^{4} + 1)}^{5}}

\frac{2 \cdot x \cdot (x^{4} + 1 - 8 x^{2} \cdot (x^{2} + 1))}{{(x^{4} + 1)}^{5}}

ⓘ Expand by multiplying each term inside the parentheses:

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

- (8 x^{4} + 8 x^{2})

2 x \cdot (x^{4} + 1 - (8 x^{4} + 8 x^{2}))

2 x^{5} + 2 x + 2 x \cdot (- (8 x^{4} + 8 x^{2}))

\frac{2 x^{5} + 2 x + 2 x \cdot (- (8 x^{4} + 8 x^{2}))}{{(x^{4} + 1)}^{5}}

ⓘ The minus sign is placed in front of the product:

2 \cdot (- (8 x^{4} + 8 x^{2}))

-2 \cdot (8 x^{4} + 8 x^{2})

\frac{2 x^{5} + 2 x - 2 x \cdot (8 x^{4} + 8 x^{2})}{{(x^{4} + 1)}^{5}}

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

\frac{2 x^{5} + 2 x - 2 x \cdot (8 x^{4} + 8 x^{2})}{{(x^{4} + 1)}^{5}}

\frac{2 x \cdot (x^{4} + 1 - (8 x^{4} + 8 x^{2}))}{{(x^{4} + 1)}^{5}}

\frac{2 x \cdot (x^{4} + 1 - (8 x^{4} + 8 x^{2}))}{{(x^{4} + 1)}^{5}}

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (8 x^{4} + 8 x^{2})

- 8 x^{4} - 8 x^{2}

\frac{2 x \cdot (x^{4} + 1 - 8 x^{4} - 8 x^{2})}{{(x^{4} + 1)}^{5}}

ⓘ Like terms are grouped by adding their coefficients:

x^{4} - 8 x^{4}

-7 x^{4}

\frac{2 x \cdot (- 7 x^{4} + 1 - 8 x^{2})}{{(x^{4} + 1)}^{5}}

5) Then you can continue the derivation:

ⅰ

\frac{d}{d x} (\frac{x^{2} + 1}{{(x^{4} + 1)}^{4}})

\frac{2 x \cdot (- 7 x^{4} + 1 - 8 x^{2})}{{(x^{4} + 1)}^{5}}

\frac{2 x \cdot (- 7 x^{4} + 1 - 8 x^{2})}{{(x^{4} + 1)}^{5}}

▶Derivative:

\frac{2 x \cdot (- 7 x^{4} - 8 x^{2} + 1)}{{(x^{4} + 1)}^{5}}