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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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It's 24k-6

By order of operations, you do the multiplication first: 3k(4) = 12k

12k-6+12k now combine like terms

24k-6

Hope that helps




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