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

We save the definition:

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

1) Let's perform the differentiation:

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

For

p1

x^{2}

and

q2

1 + 4 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 differentiate

q2

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

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

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

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

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

ⓘ Adding 0 changes nothing:

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

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

■A constant factor stays in front of the derivative, (k*u)' = k*u':

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

4

4

4) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

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

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

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

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

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

ⓘ Like terms are grouped by adding their coefficients:

4 x - 2 x

2 x

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

5) Then you can resume the derivation:

ⅰ

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

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

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

▶Derivative:

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