derivative of (x^2)/((x^2-1)^4) — step-by-step solution

1) Let's differentiate:

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

We differentiate

p1

x^{2}

and

q2

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

using:

\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^{2} - 1)}^{4})

We differentiate using:

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

(where f =

x^{2} - 1

)

ⓘ The derivative of x^n is n*x^(n-1):

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

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

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

ⓘ 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

4 \cdot (\frac{d}{d x} (x^{2}) + 0) \cdot {(x^{2} - 1)}^{3}

ⓘ 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

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

ⅰ

4 \cdot 2

8

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

4) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

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

ⓘ Multiplying powers with the same base adds the exponents:

x^{2} x

x^{3}

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

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

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

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

ⅰ

4 \cdot 2

8

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

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

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

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

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

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

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

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

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

We simplify the fraction:

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

by the common factor:

{(x^{2} - 1)}^{3}

to get the new quotient:

\frac{1}{{(x^{2} - 1)}^{5}}

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

ⓘ Like terms are grouped by adding their coefficients:

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

-3 x^{2}

\frac{2 x \cdot (- 3 x^{2} - 1)}{{(x^{2} - 1)}^{5}}

ⓘ The minus sign is factored out, changing the sign of each term:

- 3 x^{2} - 1

- (3 x^{2} + 1)

\frac{2 x \cdot (- (3 x^{2} + 1))}{{(x^{2} - 1)}^{5}}

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

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

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

- \frac{2 x \cdot (3 x^{2} + 1)}{{(x^{2} - 1)}^{5}}

ⓘ Expand by multiplying each term inside the parentheses:

2 x \cdot (3 x^{2} + 1)

6 x^{3} + 2 x

- \frac{6 x^{3} + 2 x}{{(x^{2} - 1)}^{5}}

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

- \frac{6 x^{3} + 2 x}{{(x^{2} - 1)}^{5}}

\frac{2 x \cdot (- (3 x^{2} + 1))}{{(x^{2} - 1)}^{5}}

\frac{2 x \cdot (- (3 x^{2} + 1))}{{(x^{2} - 1)}^{5}}

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

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

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

- \frac{2 x \cdot (3 x^{2} + 1)}{{(x^{2} - 1)}^{5}}

5) Then you can continue the derivation:

ⅰ

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

- \frac{2 x \cdot (3 x^{2} + 1)}{{(x^{2} - 1)}^{5}}

▶Derivative:

- \frac{2 x \cdot (3 x^{2} + 1)}{{(x^{2} - 1)}^{5}}