What is the derivative of ((t^3)/(t^8+1))^2 ?

The derivative of ((t^3)/(t^8+1))^2 is (2t^3(-5t^{10}+3t^2))/((t^8+1)^3)

What is the first derivative of ((t^3)/(t^8+1))^2 ?

The first derivative of ((t^3)/(t^8+1))^2 is (2t^3(-5t^{10}+3t^2))/((t^8+1)^3)

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derivative of

(\frac{t ^{3}}{t ^{8} + 1})^{2}

\frac{2 t ^{3} ( - 5 t ^{10} + 3 t ^{2} )}{( t ^{8} + 1 ) ^{3}}

Solution steps

\frac{d}{d t} ((\frac{t ^{3}}{t ^{8} + 1})^{2})

Apply the chain rule:

2 \cdot \frac{t ^{3}}{t ^{8} + 1} \frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 1})

= 2 \cdot \frac{t ^{3}}{t ^{8} + 1} \frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 1})

\frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 1}) = \frac{- 5 t ^{10} + 3 t ^{2}}{( t ^{8} + 1 ) ^{2}}

= 2 \cdot \frac{t ^{3}}{t ^{8} + 1} \cdot \frac{- 5 t ^{10} + 3 t ^{2}}{( t ^{8} + 1 ) ^{2}}

Simplify

2 \cdot \frac{t ^{3}}{t ^{8} + 1} \cdot \frac{- 5 t ^{10} + 3 t ^{2}}{( t ^{8} + 1 ) ^{2}} : \frac{2 t ^{3} ( - 5 t ^{10} + 3 t ^{2} )}{( t ^{8} + 1 ) ^{3}}

= \frac{2 t ^{3} ( - 5 t ^{10} + 3 t ^{2} )}{( t ^{8} + 1 ) ^{3}}

y^{^{''}} + 4 y = cos (t)

x \to 6 lim (8 x^{2} - 3 x + 1)

x \to - 5 + lim (\frac{5 x}{x + 5})

f (x) = 2 x^{2} - 9 x + 10

What is the derivative of ((t^3)/(t^8+1))^2 ?
The derivative of ((t^3)/(t^8+1))^2 is (2t^3(-5t^{10}+3t^2))/((t^8+1)^3)
What is the first derivative of ((t^3)/(t^8+1))^2 ?
The first derivative of ((t^3)/(t^8+1))^2 is (2t^3(-5t^{10}+3t^2))/((t^8+1)^3)