What is the derivative of ((t^3)/(t^6+9))^2 ?

The derivative of ((t^3)/(t^6+9))^2 is (2t^3(-3t^8+27t^2))/((t^6+9)^3)

What is the first derivative of ((t^3)/(t^6+9))^2 ?

The first derivative of ((t^3)/(t^6+9))^2 is (2t^3(-3t^8+27t^2))/((t^6+9)^3)

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derivative of

(\frac{t ^{3}}{t ^{6} + 9})^{2}

\frac{2 t ^{3} ( - 3 t ^{8} + 27 t ^{2} )}{( t ^{6} + 9 ) ^{3}}

Solution steps

\frac{d}{d t} ((\frac{t ^{3}}{t ^{6} + 9})^{2})

Apply the chain rule:

2 \cdot \frac{t ^{3}}{t ^{6} + 9} \frac{d}{d t} (\frac{t ^{3}}{t ^{6} + 9})

= 2 \cdot \frac{t ^{3}}{t ^{6} + 9} \frac{d}{d t} (\frac{t ^{3}}{t ^{6} + 9})

\frac{d}{d t} (\frac{t ^{3}}{t ^{6} + 9}) = \frac{- 3 t ^{8} + 27 t ^{2}}{( t ^{6} + 9 ) ^{2}}

= 2 \cdot \frac{t ^{3}}{t ^{6} + 9} \cdot \frac{- 3 t ^{8} + 27 t ^{2}}{( t ^{6} + 9 ) ^{2}}

Simplify

2 \cdot \frac{t ^{3}}{t ^{6} + 9} \cdot \frac{- 3 t ^{8} + 27 t ^{2}}{( t ^{6} + 9 ) ^{2}} : \frac{2 t ^{3} ( - 3 t ^{8} + 27 t ^{2} )}{( t ^{6} + 9 ) ^{3}}

= \frac{2 t ^{3} ( - 3 t ^{8} + 27 t ^{2} )}{( t ^{6} + 9 ) ^{3}}

5 e^{x^{2}}

(- 2, 5), (3, 5)

y = e^{8 e^{x}}

\frac{d}{d x} (\frac{1}{8} x^{4} + \frac{1}{4} x^{- 2})

y^{^{''}} + 25 y = cos (3 t)

What is the derivative of ((t^3)/(t^6+9))^2 ?
The derivative of ((t^3)/(t^6+9))^2 is (2t^3(-3t^8+27t^2))/((t^6+9)^3)
What is the first derivative of ((t^3)/(t^6+9))^2 ?
The first derivative of ((t^3)/(t^6+9))^2 is (2t^3(-3t^8+27t^2))/((t^6+9)^3)