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Katarina [22]
2 years ago
13

54. Find

ddle" class="latex-formula"> if x^y=y^x\text{.}
Mathematics
1 answer:
Stella [2.4K]2 years ago
7 0

Step-by-step explanation:

Take the natural log of both sides:

ln ({x}^{y} ) = ln ({y}^{x} )

Logarithm rules allow you to bring down the exponents:

yln(x) = xln(y)

Now differentiate. We will have to implicitly differentiate 'y' since it is a function of 'x'. Both sides require the product rule:

\frac{dy}{dx} ln(x) +  \frac{y}{x}  = ln(y) +  \frac{x}{y}  \frac{dy}{dx}

Isolate the terms that have y' since that is what we want:

\frac{dy}{dx} ln(x) -  \frac{x}{y}  \frac{dy}{dx}= ln(y) - \frac{y}{x}

Factor out y' to get:

\frac{dy}{dx}( ln(x) -   \frac{x}{y})= ln(y) - \frac{y}{x}

Therefore:

\frac{dy}{dx} =  \frac{ln(y) -  \frac{y}{x} }{ln(x) -  \frac{x}{y} }

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This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
noname [10]

L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z

5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

3 0
3 years ago
PLZZZZ HELP THIS IS WORTH 40 POINTS PLZZZ<br> I’ll give extra points!!!
iVinArrow [24]

Answer:

1) 15

2) 10

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Step-by-step explanation:

Substitute the values thats all lol, if u need a more detailed explanation just comment <3

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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