What is the derivative of (1-2t)^{-4}(t^2-t)^2 ?

The derivative of (1-2t)^{-4}(t^2-t)^2 is -(2t(t-1))/((-2t+1)^5)

What is the first derivative of (1-2t)^{-4}(t^2-t)^2 ?

The first derivative of (1-2t)^{-4}(t^2-t)^2 is -(2t(t-1))/((-2t+1)^5)

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derivative of

(1 - 2 t)^{- 4} (t^{2} - t)^{2}

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (1 - 2 t)^{- 4}, g = (t^{2} - t)^{2}

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

\frac{d}{d t} ((1 - 2 t)^{- 4}) = \frac{8}{( 1 - 2 t ) ^{5}}

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

= \frac{8}{( 1 - 2 t ) ^{5}} (t^{2} - t)^{2} + 2 (t^{2} - t) (2 t - 1) (1 - 2 t)^{- 4}

Simplify

\frac{8}{( 1 - 2 t ) ^{5}} (t^{2} - t)^{2} + 2 (t^{2} - t) (2 t - 1) (1 - 2 t)^{- 4} : - \frac{2 t ( t - 1 )}{( - 2 t + 1 ) ^{5}}

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

\frac{\partial}{\partial y} (2 x^{\frac{y}{z}})

\int \frac{( 1 - x ) ^{2}}{x ^{2}} d x

\frac{d}{d x} (\frac{1}{2} ln^{2} (x))

- 7 csc (5 x) e^{3 x}

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

What is the derivative of (1-2t)^{-4}(t^2-t)^2 ?
The derivative of (1-2t)^{-4}(t^2-t)^2 is -(2t(t-1))/((-2t+1)^5)
What is the first derivative of (1-2t)^{-4}(t^2-t)^2 ?
The first derivative of (1-2t)^{-4}(t^2-t)^2 is -(2t(t-1))/((-2t+1)^5)