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alexira [117]
3 years ago
7

Please Help this is Due in 1 hour and I really dont understand any of this. (Question in Image)

Mathematics
1 answer:
seropon [69]3 years ago
4 0

Answer:

Step-by-step explanation:

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An integer is always a whole number

Step-by-step explanation:

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Find dy/dx if y =x^3+5x+2/x²-1
stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

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  • g(x) = x^2-1

Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

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