What is the derivative of (t^6)/((t^3+1)^2) ?

The derivative of (t^6)/((t^3+1)^2) is (6t^5)/((t^3+1)^3)

What is the first derivative of (t^6)/((t^3+1)^2) ?

The first derivative of (t^6)/((t^3+1)^2) is (6t^5)/((t^3+1)^3)

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derivative of

\frac{t ^{6}}{( t ^{3} + 1 ) ^{2}}

\frac{6 t ^{5}}{( t ^{3} + 1 ) ^{3}}

Solution steps

\frac{d}{d t} (\frac{t ^{6}}{( t ^{3} + 1 ) ^{2}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d t} ( t ^{6} ) ( t ^{3} + 1 ) ^{2} - \frac{d}{d t} ( ( t ^{3} + 1 ) ^{2} ) t ^{6}}{( ( t ^{3} + 1 ) ^{2} ) ^{2}}

\frac{d}{d t} (t^{6}) = 6 t^{5}

\frac{d}{d t} ((t^{3} + 1)^{2}) = 6 t^{2} (t^{3} + 1)

= \frac{6 t ^{5} ( t ^{3} + 1 ) ^{2} - 6 t ^{2} ( t ^{3} + 1 ) t ^{6}}{( ( t ^{3} + 1 ) ^{2} ) ^{2}}

Simplify

\frac{6 t ^{5} ( t ^{3} + 1 ) ^{2} - 6 t ^{2} ( t ^{3} + 1 ) t ^{6}}{( ( t ^{3} + 1 ) ^{2} ) ^{2}} : \frac{6 t ^{5}}{( t ^{3} + 1 ) ^{3}}

= \frac{6 t ^{5}}{( t ^{3} + 1 ) ^{3}}

\int 6 tan^{5} (x) sec^{2} (x) d x

x \to 0 - lim (\frac{8}{5 x ^{\frac{1}{5}}})

\int \frac{x ^{2}}{x - 5} d x

\frac{d y}{d x} = 0.2 \cdot y + 200

(x^{3} + y^{3}) d x - x y^{2} d y = 0

What is the derivative of (t^6)/((t^3+1)^2) ?
The derivative of (t^6)/((t^3+1)^2) is (6t^5)/((t^3+1)^3)
What is the first derivative of (t^6)/((t^3+1)^2) ?
The first derivative of (t^6)/((t^3+1)^2) is (6t^5)/((t^3+1)^3)