derivative of -2/(9(1+x)^{5/3}) — step-by-step solution

⚠ -2/(9(1+x)^{5/3}) is read as -2/(9(1+x)^(5/3)): use ^(...) to group an exponent, use * to multiply

To ease differentiation, the expression

- \frac{2}{9 \cdot {(1 + x)}^{\frac{5}{3}}}

is turned into

- \frac{2}{9} \cdot {(1 + x)}^{- \frac{5}{3}}

Let's perform the differentiation:

\frac{d}{d x} (- \frac{2}{9} \cdot {(1 + x)}^{- \frac{5}{3}})

We differentiate using:

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

(where f =

1 + x

)

ⓘ Multiplying a fraction by a term multiplies its numerator:

- \frac{2}{9} \cdot {(1 + x)}^{- \frac{5}{3}}

- \frac{2 \cdot {(1 + x)}^{- \frac{5}{3}}}{9}

■A constant factor stays in front of the derivative, (k*u)' = k*u'. The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (- \frac{2 \cdot {(1 + x)}^{- \frac{5}{3}}}{9})

- \frac{2}{9} \cdot (- \frac{5}{3} \cdot \frac{d}{d x} (1 + x) \cdot {(1 + x)}^{- \frac{8}{3}})

- \frac{2}{9} \cdot (- \frac{5}{3} \cdot \frac{d}{d x} (1 + x) \cdot {(1 + x)}^{- \frac{8}{3}})

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

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

0 + 1

- \frac{2}{9} \cdot (- \frac{5}{3}) \cdot (0 + 1) \cdot {(1 + x)}^{- \frac{8}{3}}

ⓘ Adding 0 changes nothing:

0 + 1

1

- \frac{2}{9} \cdot (- \frac{5}{3}) \cdot {(1 + x)}^{- \frac{8}{3}}

ⓘ To multiply fractions, multiply the numerators together and the denominators together:

- \frac{2}{9} \cdot (- \frac{5}{3})

\frac{10}{27}

\frac{10}{27} \cdot {(1 + x)}^{- \frac{8}{3}}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{10}{27} \cdot {(1 + x)}^{- \frac{8}{3}}

\frac{10 \cdot {(1 + x)}^{- \frac{8}{3}}}{27}

■The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

\frac{10 \cdot {(1 + x)}^{- \frac{8}{3}}}{27}

\frac{10}{27 \cdot {(1 + x)}^{\frac{8}{3}}}

\frac{10}{27 \cdot {(1 + x)}^{\frac{8}{3}}}

▶Derivative:

\frac{10}{27 \cdot {(x + 1)}^{\frac{8}{3}}}