Question

if x^y ×y^x =c,then dy/dx at (e.e) is​

Answer 1

Step-by-step explanation:

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Answer 2

To find the derivative dy/dx of y with respect to x at the point (e, e), we need to use logarithmic differentiation.

Taking the natural logarithm of both sides of the given equation, we get:

ln(x^y * y^x) = ln(c)

Using the product rule of logarithms, we can simplify this expression:

ln(x^y) + ln(y^x) = ln(c)

Using the power rule of logarithms, we can further simplify:

y ln(x) + x ln(y) = ln(c)

Now we differentiate both sides of this equation with respect to x, using the product rule:

y (1/x) + x (dy/dx) (1/y) = 0

Simplifying and solving for dy/dx, we get:

dy/dx = -y/x * (1/ln(y) + x/y * 1/ln(x))

Now we can substitute x = e and y = e, since we want to find the derivative at the point (e, e):

dy/dx = -e/e * (1/ln(e) + e/e * 1/ln(e))

dy/dx = -1/ln(e) - 1/ln(e)

dy/dx = -2

Therefore, the derivative of y with respect to x at the point (e, e) is -2.

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