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d x-\left(x \ln y-y^{y}\right) d y=0

Jan 30, 2024

d x-\left(x \ln y-y^{y}\right) d y=0

Solution by Steps

step 1

Differentiate the equation

x \ln(y) - y^y = 0

implicitly with respect to

x

step 2

Applying the product rule to

x \ln(y)

, we get

\ln(y) + \frac{x}{y}y'

step 3

Differentiating

y^y

with respect to

x

gives

y^y \ln(y) y'

step 4

Set the derivative of the left side equal to the derivative of the right side (which is 0) to get

\ln(y) + \frac{x}{y}y' - y^y \ln(y) y' = 0

step 5

Solve for

y'

to find

y' = \frac{y \ln(y)}{-x + y^{y+1} + y^{y+1} \ln(y)}

step 6

Simplify the expression for

y'

to get

y' = -\frac{y \ln(y)}{x - y^{y+1} (\ln(y) + 1)}

step 7

Further simplify to

y' = \frac{y \ln(y)}{y^{y+1} (\ln(y) + 1) - x}

step 8

The final simplified form is

y' = \frac{y \ln(y)}{-x + y^{1+y} + y^{1+y} \ln(y)}

Answer

y' = \frac{y \ln(y)}{-x + y^{1+y} + y^{1+y} \ln(y)}

Key Concept

Implicit Differentiation

Explanation

When differentiating an equation implicitly, each term is differentiated with respect to

x

, and the chain rule is applied to functions of

y

that are treated as functions of

x

through

y(x)

. This often involves finding

y'

, the derivative of

y

with respect to

x