find the derivative x^8-1/x^4-1 with respect to x

Subject:
Math
Author:
lancesingh
Created:
1 year ago

Answers 2

Answer:

final answer is x⁴-1

Step-by-step explanation:

step1: (x-1)^8-4

step2: (x-1)^4

step3: x^4-1^4

step4: x⁴-1

Author:

herculescasey
Rate an answer:
10

Solution:

Given to evaluate:

[tex] \tt = \dfrac{d}{dx} \bigg( \dfrac{ {x}^{8} - 1}{ {x}^{4} - 1} \bigg)[/tex]

Can be written as:

[tex] \tt = \dfrac{d}{dx}\dfrac{({x}^{4})^{2} - {(1)}^{2} }{ {x}^{4} - 1}[/tex]

[tex] \tt = \dfrac{d}{dx}\dfrac{({x}^{4} + 1)( {x}^{4} - 1)}{ {x}^{4} - 1}[/tex]

[tex] \tt = \dfrac{d}{dx}({x}^{4} + 1)[/tex]

We know that:

[tex] \tt \longrightarrow \dfrac{d}{dx}(f \pm g) = \dfrac{df}{dx} \pm \dfrac{dg}{dx} [/tex]

Using this result, we get:

[tex] \tt = \dfrac{d}{dx}({x}^{4})+ \dfrac{d}{dx}(1)[/tex]

We know that:

[tex] \tt \longrightarrow \dfrac{d}{dx}( {x}^{n} ) =n {x}^{n - 1} [/tex]

[tex] \tt \longrightarrow \dfrac{d}{dx}(k) =0[/tex]

Using this result, we get:

[tex] \tt = 4{x}^{3} + 0[/tex]

[tex] \tt = 4{x}^{3} [/tex]

Therefore:

[tex] \tt \longrightarrow\dfrac{d}{dx} \bigg( \dfrac{ {x}^{8} - 1}{ {x}^{4} - 1} \bigg) =4 {x}^{3} [/tex]

Which is our required answer.

Learn More:

[tex]\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}[/tex]

Author:

dannaoujz
Rate an answer:
1

find the derivative x^8-1/x^4-1 with respect to x

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find the derivative x^8-1/x^4-1 with respect to x