What is the d/(dt)(((2t+5)/(t^2+1))^4) ?

The d/(dt)(((2t+5)/(t^2+1))^4) is (4(2t+5)^3(-2t^2-10t+2))/((t^2+1)^5)

What is the first d/(dt)(((2t+5)/(t^2+1))^4) ?

The first d/(dt)(((2t+5)/(t^2+1))^4) is (4(2t+5)^3(-2t^2-10t+2))/((t^2+1)^5)

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\frac{d}{d t} ((\frac{2 t + 5}{t ^{2} + 1})^{4})

\frac{4 ( 2 t + 5 ) ^{3} ( - 2 t ^{2} - 10 t + 2 )}{( t ^{2} + 1 ) ^{5}}

Solution steps

\frac{d}{d t} ((\frac{2 t + 5}{t ^{2} + 1})^{4})

Apply the chain rule:

4 (\frac{2 t + 5}{t ^{2} + 1})^{3} \frac{d}{d t} (\frac{2 t + 5}{t ^{2} + 1})

= 4 (\frac{2 t + 5}{t ^{2} + 1})^{3} \frac{d}{d t} (\frac{2 t + 5}{t ^{2} + 1})

\frac{d}{d t} (\frac{2 t + 5}{t ^{2} + 1}) = \frac{- 2 t ^{2} - 10 t + 2}{( t ^{2} + 1 ) ^{2}}

= 4 (\frac{2 t + 5}{t ^{2} + 1})^{3} \frac{- 2 t ^{2} - 10 t + 2}{( t ^{2} + 1 ) ^{2}}

Simplify

4 (\frac{2 t + 5}{t ^{2} + 1})^{3} \frac{- 2 t ^{2} - 10 t + 2}{( t ^{2} + 1 ) ^{2}} : \frac{4 ( 2 t + 5 ) ^{3} ( - 2 t ^{2} - 10 t + 2 )}{( t ^{2} + 1 ) ^{5}}

= \frac{4 ( 2 t + 5 ) ^{3} ( - 2 t ^{2} - 10 t + 2 )}{( t ^{2} + 1 ) ^{5}}

\frac{\partial}{\partial u} (\frac{1}{u ^{2} + v ^{2}})

\frac{d}{d I} (\frac{100 I}{I ^{2} + I + 4})

\int \frac{y ^{6}}{2 y ^{7} + 2} d y

\frac{d ^{2}}{d x ^{2}} (cos (x) + sin (x) + \frac{1}{2} x^{2} + \frac{1}{x})

\frac{d}{d x} (2 (sin (x))^{2})

What is the d/(dt)(((2t+5)/(t^2+1))^4) ?
The d/(dt)(((2t+5)/(t^2+1))^4) is (4(2t+5)^3(-2t^2-10t+2))/((t^2+1)^5)
What is the first d/(dt)(((2t+5)/(t^2+1))^4) ?
The first d/(dt)(((2t+5)/(t^2+1))^4) is (4(2t+5)^3(-2t^2-10t+2))/((t^2+1)^5)