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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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\int_{- 1}^{4} x - 3 x^{2} d x

(x e^{x^{2}})^{^{'}}

\frac{\partial}{\partial x} (e^{x y - y})

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