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

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

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

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

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derivative of

f (t) = \frac{1}{2 t + 3}

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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