What is the value of d/dx (6/x^4 + 1/x^2) at x = -1 ?

Mathematics

1 answer:

zheka24 [161]2 years ago

7 0

Answer:

\frac{d}{dx}\left(\frac{1+x^4+x^6}{x^2+x+1}\right)=\frac{4x^7+5x^6+8x^5+3x^4+4x^3-2x-1}{\left(x^2+x+1\right)^2}

Step-by-step explanation:

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$(x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x$

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$\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+x\,\vec k$

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$\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k$ ,

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$=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy$