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Popular >

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

Solution

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

Solution

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

Popular Examples

tangent of f(x)=x+1-ln(x),\at x=0.5 tangent of

f (x) = x + 1 - ln (x), at x = 0.5

(t^2+x^2)dt+(tx)dx=0

(t^{2} + x^{2}) d t + (t x) d x = 0

integral from 0 to pi of 1/(sqrt(x))

\int_{0}^{π} \frac{1}{x} d x

derivative of ln(xsqrt(x^2-7))

\frac{d}{d x} (ln (x x^{2} - 7))

inverse oflaplace 1/(s^2*(s^2+4))inverse laplace

\frac{1}{s ^{2} \cdot ( s ^{2} + 4 )}

Frequently Asked Questions (FAQ)

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