What is the derivative of 3t^3-2/(t^2) ?

The derivative of 3t^3-2/(t^2) is 9t^2+4/(t^3)

What is the first derivative of 3t^3-2/(t^2) ?

The first derivative of 3t^3-2/(t^2) is 9t^2+4/(t^3)

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derivative of

3 t^{3} - \frac{2}{t ^{2}}

9 t^{2} + \frac{4}{t ^{3}}

Solution steps

\frac{d}{d t} (3 t^{3} - \frac{2}{t ^{2}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d t} (\frac{2}{t ^{2}}) = - \frac{4}{t ^{3}}

= 9 t^{2} - (- \frac{4}{t ^{3}})

Simplify

= 9 t^{2} + \frac{4}{t ^{3}}

f (x) = x^{2} - 6

\int_{0}^{\infty} 77 \frac{e ^{x}}{e ^{2 x} + 3} d x

y \cdot y^{^{'}} = 1

\int \frac{1}{1 - 4 ^{2}} d x

\int_{- 4}^{4} (16 - x^{2}) d x

What is the derivative of 3t^3-2/(t^2) ?
The derivative of 3t^3-2/(t^2) is 9t^2+4/(t^3)
What is the first derivative of 3t^3-2/(t^2) ?
The first derivative of 3t^3-2/(t^2) is 9t^2+4/(t^3)