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2 years ago

How can i differentiate this equation?

Mathematics

1 answer:

Dmitry_Shevchenko [17]2 years ago

8 0

$\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}$

$\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}$

$\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}$

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Which ordered pair is a solution to the system of equations?

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Answer:

See below

Step-by-step explanation:

Given system:

2x - y = - 2
15x - 2y = - 7

Solve by elimination, multiply the first equation by - 2 and add up the equations:

-2(2x - y) + 15x - 2y = - 2(- 2) - 7
-4x + 2y + 15x - 2y = 4 - 7
11x = - 3
x = - 3/11

Find the value of y:

2( - 3/11) - y = - 2
-6/11 - y = - 2
y = 2 - 6/11
y = 1/11

Solution is:

( - 3/11, 1/11)

None of the answer choices is matching this solution. Something must be wrong with the given equations.

3 0

2 years ago

Ayuda con mates por favor

QveST [7]

Answer:

a) $x = -1$ and $y = 5$ , b) $x = 2$ and $y = -1$ , c) $x = -2$ and $y = 1$ , d) $x = 2$ and $y = -9$ , e) $x = 2$ and $y = -2$ , f) $x = 2$ and $y = -1$ , g) $x = 2$ and $y = -1$ , h) $x = 1$ and $y = 2$ .

Step-by-step explanation:

Each system is solved as follows (Cada sistema es resuelto como sigue):

a) $2\cdot x + y = 3$ and $3\cdot x - y = -8$

$y = 3 - 2\cdot x$

$3\cdot x - 3 + 2\cdot x = -8$

$5\cdot x = -5$

$x = -1$

$y = 5$

b) $x - 2\cdot y = 4$ and $-x+3\cdot y = -5$

$x = 4 + 2\cdot y$

$-4-2\cdot y+3\cdot y = - 5$

$-4+y = -5$

$y = -1$

$x = 2$

c) $-2\cdot x + 5\cdot y = 9$ and $x - y = -3$

$x = y-3$

$-2\cdot y +6 +5\cdot y = 9$

$3\cdot y = 3$

$y = 1$

$x = -2$

d) $5\cdot x - 4\cdot y = 2$ and $3\cdot x + 2\cdot y = -12$

$3\cdot x + 2\cdot y = -12$

$6\cdot x + 4\cdot y = -24$

$4\cdot y = -6\cdot x -24$

$5\cdot x +6\cdot x +24 = 2$

$11\cdot x = -22$

$x = 2$

$y = -9$

e) $x + y = 0$ and $2\cdot x +3\cdot y = -2$

$y = -x$

$2\cdot x -3\cdot x = -2$

$-x = -2$

$x = 2$

$y = -2$

f) $3\cdot x + 5\cdot y = 1$ and $x + y = 1$

$y = 1 - x$

$3\cdot x + 5 - 5\cdot x = 1$

$-2\cdot x = -4$

$x = 2$

$y = -1$

g) $5\cdot x - 2\cdot y = 12$ and $4\cdot x + 3\cdot y = 5$

$x = \frac{12+2\cdot y}{5}$

$x = \frac{5-3\cdot y}{4}$

$\frac{12+2\cdot y}{5} = \frac{5-3\cdot y}{4}$

$4\cdot (12+2\cdot y) = 5\cdot (5-3\cdot y)$

$48 + 8\cdot y = 25 - 15\cdot y$

$23\cdot y = -23$

$y = -1$

$x = 2$

h) $7\cdot x + 8\cdot y = 23$ and $3\cdot x + 2\cdot y = 7$

$3\cdot x + 2\cdot y = 7$

$12\cdot x + 8\cdot y = 28$

$8\cdot y = 28 - 12\cdot x$

$7\cdot x +28-12\cdot x = 23$

$-5\cdot x = -5$

$x = 1$

$y = 2$

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3 years ago

If ΔΡQR

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Answer:

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Step-by-step explanation:

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2 years ago

-6 is less than or equal to x, and -1 is greater than x

Olenka [21]

Answer:

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