derivative of f(x)=(x^2)/(2+x) — step-by-step solution

We save the definition:

f (x) = \frac{x^{2}}{2 + x}

1) Let's perform the differentiation:

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

For

p1

x^{2}

and

q2

2 + x

, 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} (2 + x)

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

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

0 + 1

0 + 1

ⓘ Adding 0 changes nothing:

0 + 1

1

1

4) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

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

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

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

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

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

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

ⓘ Expand by multiplying each term inside the parentheses:

2 \cdot (x + 2)

2 x + 4

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

ⓘ Like terms are grouped by adding their coefficients:

2 x - x

x

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

5) You can then resume the derivation:

ⅰ

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

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

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

▶Derivative:

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