derivative of (0.2x/(x^2+9)) — step-by-step solution

1) Let's perform the differentiation:

\frac{d}{d x} (\frac{0.2 x}{x^{2} + 9})

We differentiate

p1

0.2 x

and

q2

x^{2} + 9

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} (0.2 x)

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

\frac{d}{d x} (0.2 x)

0.2

0.2

3) Let's perform the differentiation

q2

\frac{d}{d x} (x^{2} + 9)

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

\frac{d}{d x} (x^{2} + 9)

\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

4) Let's evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{0.2 \cdot (x^{2} + 9) - 0.2 x \cdot 2 x}{{(x^{2} + 9)}^{2}}

ⓘ Multiplications are done before additions and subtractions:

0.2 \cdot 2

0.2 + 0.2

0.4

■Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

\frac{0.2 \cdot (x^{2} + 9) - 0.4 x^{2}}{{(x^{2} + 9)}^{2}}

ⓘ Expand by multiplying each term inside the parentheses:

0.2 \cdot (x^{2} + 9)

0.2 x^{2} + 1.8

\frac{0.2 x^{2} + 1.8 - 0.4 x^{2}}{{(x^{2} + 9)}^{2}}

ⓘ Like terms are grouped by adding their coefficients:

0.2 x^{2} - 0.4 x^{2}

-0.2 x^{2}

\frac{- 0.2 x^{2} + 1.8}{{(x^{2} + 9)}^{2}}

5) You can then resume the differentiation:

ⅰ

\frac{d}{d x} (\frac{0.2 x}{x^{2} + 9})

\frac{- 0.2 x^{2} + 1.8}{{(x^{2} + 9)}^{2}}

\frac{- 0.2 x^{2} + 1.8}{{(x^{2} + 9)}^{2}}

▶Derivative:

\frac{1.8 - 0.2 x^{2}}{{(x^{2} + 9)}^{2}}