What is the derivative of 4/((x^3+9^5)) ?

The derivative of 4/((x^3+9^5)) is -(60x^2)/((x^3+9)^6)

What is the first derivative of 4/((x^3+9^5)) ?

The first derivative of 4/((x^3+9^5)) is -(60x^2)/((x^3+9)^6)

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\frac{d}{d x} (\frac{4}{( x ^{3} + 9 ) ^{5}})

- \frac{60 x ^{2}}{( x ^{3} + 9 ) ^{6}}

Solution steps

\frac{d}{d x} (\frac{4}{( x ^{3} + 9 ) ^{5}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} (\frac{1}{( x ^{3} + 9 ) ^{5}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 4 \frac{d}{d x} ((x^{3} + 9)^{- 5})

Apply the chain rule:

- \frac{5}{( x ^{3} + 9 ) ^{6}} \frac{d}{d x} (x^{3} + 9)

= - \frac{5}{( x ^{3} + 9 ) ^{6}} \frac{d}{d x} (x^{3} + 9)

\frac{d}{d x} (x^{3} + 9) = 3 x^{2}

= 4 (- \frac{5}{( x ^{3} + 9 ) ^{6}} \cdot 3 x^{2})

Simplify

4 (- \frac{5}{( x ^{3} + 9 ) ^{6}} \cdot 3 x^{2}) : - \frac{60 x ^{2}}{( x ^{3} + 9 ) ^{6}}

= - \frac{60 x ^{2}}{( x ^{3} + 9 ) ^{6}}

\frac{d x}{d t} + 2 t^{5} x^{7} + \frac{x}{t} = 0

n = 0 \sum \infty e^{- 6 n}

(7, - 4), (8, 8)

\frac{d}{d x} (\frac{- ln ( 1 + 6 x )}{x})

\int 1 - 2 y d y

What is the derivative of 4/((x^3+9^5)) ?
The derivative of 4/((x^3+9^5)) is -(60x^2)/((x^3+9)^6)
What is the first derivative of 4/((x^3+9^5)) ?
The first derivative of 4/((x^3+9^5)) is -(60x^2)/((x^3+9)^6)