Mathematics, 21.07.2019 21:40, 09daishagreen

Use the chain rule to find dz/dt. z = xy9 − x2y, x = t2 + 1, y = t2 − 1

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answered: rburd

The value of dz/dt using the chain rule is 2t[(t^2-1)^9 -2(t^4-1)]+2t[9(t^2+1)(t^2-1)^8-(t^2+1)^2]

According to chain rule:

$\frac{dz}{dt} = \frac{\delta z}{\delta x} \frac{dx}{dt} + \frac{\delta z}{\delta y} \frac{dy}{dt}$

Given the following expressions

z = xy⁹ − x²y,

x = t² + 1

y = t² − 1

Get the differentials in the equation

$\delta z/\delta x = y^9-2xy\\dzx/dt = 2t\\\delta z/\delta y =9xy^8-x^2\\dy/dt = 2t$

Substitute the differentials into the formula

$\frac{dz}{dt} = (y^9-2xy)(2t)+(9xy^8-x^2)(2t)\\\frac{dz}{dt} = 2t[(t^2-1)^9 -2(t^4-1)]+2t[9(t^2+1)(t^2-1)^8-(t^2+1)^2]$

Hence the value of dz/dt using the chain rule is 2t[(t^2-1)^9 -2(t^4-1)]+2t[9(t^2+1)(t^2-1)^8-(t^2+1)^2]

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answered: nmartin5185

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial 
x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial 
y}\dfrac{\mathrm dy}{\mathrm dt}$
$\dfrac{\mathrm dz}{\mathrm dt}=(y^9-2xy)(2t)+(9xy^8-x^2)(2t)$
$\dfrac{\mathrm dz}{\mathrm dt}=20t^{19}-144t^{17}+432t^{15}-672t^{13}+504t^{11}-336t^7+282t^5-112t^3+18t$

answered: Guest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

answered: Guest

the length of o'p' is $1.5\times op$ .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

$k=\frac{o'p'}{op}$

the scale factor is 1.5.

$1.5=\frac{o'p'}{op}$

multiply both sides by op.

$1.5\times op=o'p'$

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is $1.5\times op$ .

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Use the chain rule to find dz/dt. z = xy9 − x2y, x = t2 + 1, y = t2 − 1...