What is the derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) ?

The derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) is (14)/(x^3)-(16)/(x^2)-(24)/(x^4)+(16)/(x^5)

What is the first derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) ?

The first derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) is (14)/(x^3)-(16)/(x^2)-(24)/(x^4)+(16)/(x^5)

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derivative of

y = (x^{2} + 8 x - 4) (2 x^{- 2} + x^{- 4})

\frac{14}{x ^{3}} - \frac{16}{x ^{2}} - \frac{24}{x ^{4}} + \frac{16}{x ^{5}}

Solution steps

\frac{d}{d x} ((x^{2} + 8 x - 4) (2 x^{- 2} + x^{- 4}))

Apply the Product Rule:

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

f = x^{2} + 8 x - 4, g = 2 x^{- 2} + x^{- 4}

= \frac{d}{d x} (x^{2} + 8 x - 4) (2 x^{- 2} + x^{- 4}) + \frac{d}{d x} (2 x^{- 2} + x^{- 4}) (x^{2} + 8 x - 4)

\frac{d}{d x} (x^{2} + 8 x - 4) = 2 x + 8

\frac{d}{d x} (2 x^{- 2} + x^{- 4}) = - \frac{4}{x ^{3}} - \frac{4}{x ^{5}}

= (2 x + 8) (2 x^{- 2} + x^{- 4}) + (- \frac{4}{x ^{3}} - \frac{4}{x ^{5}}) (x^{2} + 8 x - 4)

Simplify

(2 x + 8) (2 x^{- 2} + x^{- 4}) + (- \frac{4}{x ^{3}} - \frac{4}{x ^{5}}) (x^{2} + 8 x - 4) : \frac{14}{x ^{3}} - \frac{16}{x ^{2}} - \frac{24}{x ^{4}} + \frac{16}{x ^{5}}

= \frac{14}{x ^{3}} - \frac{16}{x ^{2}} - \frac{24}{x ^{4}} + \frac{16}{x ^{5}}

cos (4 t)

x \to 1 - lim (\frac{x - 1}{x + 1})

f (x) = \frac{x}{10}

\frac{d y}{d x}, x = y^{10}

\frac{\partial}{\partial x} (- \frac{x}{1 - x ^{2} y ^{2}})

What is the derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) ?
The derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) is (14)/(x^3)-(16)/(x^2)-(24)/(x^4)+(16)/(x^5)
What is the first derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) ?
The first derivative of y=(x^2+8x-4)(2x^{-2}+x^{-4}) is (14)/(x^3)-(16)/(x^2)-(24)/(x^4)+(16)/(x^5)