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1 year ago

8. (02.02 LC)

Mathematics

1 answer:

RSB [31]1 year ago

3 0

Answer:

$5 \sqrt{3} - 20 \sqrt{2}$

Step-by-step explanation:

$simpify \\ \sqrt{75} - 4 \sqrt{8} - 3 \sqrt{32} \\ from \: the \: above \: get \: two \: numbers \: \\ such \: that \: when \: multiplied \: gives \\ \: the \: numbers \: inside \: the \: root \\ and \: one \: of \: the \: numers \: should \\ be \: a \: perfect \: square. \\ \sqrt{25 \times 3} - 4 \sqrt{4 \times 2} - 3\sqrt{16 \times 2} \\ 5 \sqrt{3} - 4 \times 2 \sqrt{2} -3 \times 4\sqrt{2} \\ 5 \sqrt{3} - 8 \sqrt{2} - 12\sqrt{2} \\ 5 \sqrt{3} - 20 \sqrt{2}$

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3 years ago

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2 years ago

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

Differentiate using the Quotient Rule –

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

According to the given question, we have –

f(x) = x^3+5x+2
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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1 year ago