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Popular >

d/(dt)(1/(2t^2+1))

Solution

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

Solution

- \frac{4 t}{( 2 t ^{2} + 1 ) ^{2}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d t} (2 t^{2} + 1) = 4 t

= - \frac{1}{( 2 t ^{2} + 1 ) ^{2}} \cdot 4 t

Simplify

- \frac{1}{( 2 t ^{2} + 1 ) ^{2}} \cdot 4 t : - \frac{4 t}{( 2 t ^{2} + 1 ) ^{2}}

= - \frac{4 t}{( 2 t ^{2} + 1 ) ^{2}}

Graph

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Frequently Asked Questions (FAQ)

What is the d/(dt)(1/(2t^2+1)) ?
The d/(dt)(1/(2t^2+1)) is -(4t)/((2t^2+1)^2)
What is the first d/(dt)(1/(2t^2+1)) ?
The first d/(dt)(1/(2t^2+1)) is -(4t)/((2t^2+1)^2)