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

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

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

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

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

- \frac{1}{4 x ^{5}}

Solution steps

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

Treat

t

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} ((2 t^{- 1})^{2}) = 0

= - \frac{1}{4 x ^{5}} - 0

Simplify

= - \frac{1}{4 x ^{5}}

(4 x^{2} - 5 x) e^{x}

7 - x^{2}, x^{2} - 1

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

\frac{d}{d x} (x x^{2} + y^{2})

\frac{d}{d x} (\frac{2 x + 5}{x})

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