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3 years ago

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Find y prime if arctan(xy) = 1 + (x^2*y)

1 answer:

steposvetlana [31]3 years ago

3 0

I'm pretty sure that these tips which I give you will make you able to sort things out.
According to the fact that <span>derivative of the function f(x) = y, you an use these formulae in order to calculate the needed result : </span> $arctan(xy) = 1+ x^2*y =\ \textgreater \ xy = tan(1+x^2y)$
$y + xy' = sec^2(1+x^2y)(2x*y + y'*x^2)$
Do hope this will lead you to the corect answer! Regards.

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Naily [24]

Answer: z^2-4
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EleoNora [17]

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3 years ago

Kay cuts three (3) pizzas so each of her friend at the table receives ¾ pizza. How many friends are at the table?

jarptica [38.1K]

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2 years ago

The temperature function (in degrees Fahrenheit) in a three dimensional space is given by T(x, y, z) = 3x + 6y - 6z + 1. A bee i

madam [21]

You're looking for the extreme values of $T(x,y,z)=3x+6y-6z+1$ subject to $x^2+y^2+z^2=9$ . The Lagrangian is

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with critical wherever the partial derivatives vanish:

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$L_y=6+2\lambda y=0\implies y=-\dfrac3\lambda$

$L_z=-6+2\lambda z=0\implies z=\dfrac3\lambda$

$L_\lambda=x^2+y^2+z^2-9=0$

Substituting the first three solutions into the last equation gives

$\dfrac9{4\lambda^2}+\dfrac9{\lambda^2}+\dfrac9{\lambda^2}=9$

$\implies\lambda=\pm\dfrac32$

$\implies x=1,y=2,z=-2\text{ or }x=-1,y=-2,z=2$

At these points, we have

$T(1,2,-2)=28$

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so the highest temperature the bee can experience is 28º F at the point (1, 2, -2), and the lowest is -26º F at the point (-1, -2, 2).

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3 years ago

Elise bought a dress that was discounted 35% off of

bixtya [17]

Answer:

a: $49

b: $91

Explaination:

a: 35% of 140 = 49

b: 140 - 49 = 91

You're welcome! <3

6 0

2 years ago