What is the derivative of h(t)=((t^3)/(t^8+9))^2 ?

The derivative of h(t)=((t^3)/(t^8+9))^2 is (2t^3(-5t^{10}+27t^2))/((t^8+9)^3)

What is the first derivative of h(t)=((t^3)/(t^8+9))^2 ?

The first derivative of h(t)=((t^3)/(t^8+9))^2 is (2t^3(-5t^{10}+27t^2))/((t^8+9)^3)

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derivative of

h (t) = (\frac{t ^{3}}{t ^{8} + 9})^{2}

\frac{2 t ^{3} ( - 5 t ^{10} + 27 t ^{2} )}{( t ^{8} + 9 ) ^{3}}

Solution steps

\frac{d}{d t} ((\frac{t ^{3}}{t ^{8} + 9})^{2})

Apply the chain rule:

2 \cdot \frac{t ^{3}}{t ^{8} + 9} \frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 9})

= 2 \cdot \frac{t ^{3}}{t ^{8} + 9} \frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 9})

\frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 9}) = \frac{- 5 t ^{10} + 27 t ^{2}}{( t ^{8} + 9 ) ^{2}}

= 2 \cdot \frac{t ^{3}}{t ^{8} + 9} \cdot \frac{- 5 t ^{10} + 27 t ^{2}}{( t ^{8} + 9 ) ^{2}}

Simplify

2 \cdot \frac{t ^{3}}{t ^{8} + 9} \cdot \frac{- 5 t ^{10} + 27 t ^{2}}{( t ^{8} + 9 ) ^{2}} : \frac{2 t ^{3} ( - 5 t ^{10} + 27 t ^{2} )}{( t ^{8} + 9 ) ^{3}}

= \frac{2 t ^{3} ( - 5 t ^{10} + 27 t ^{2} )}{( t ^{8} + 9 ) ^{3}}

What is the derivative of h(t)=((t^3)/(t^8+9))^2 ?
The derivative of h(t)=((t^3)/(t^8+9))^2 is (2t^3(-5t^{10}+27t^2))/((t^8+9)^3)
What is the first derivative of h(t)=((t^3)/(t^8+9))^2 ?
The first derivative of h(t)=((t^3)/(t^8+9))^2 is (2t^3(-5t^{10}+27t^2))/((t^8+9)^3)