find y'(x) if x^y=y^x
x^y=y^x --->
y log(x) = x log(y) ----->
log(x) dy + y/x dx =
log(y) dx + x/y dy ---->
[log(x) - x/y] dy = [log(y) - y/x] dx
---->
dy/dx = [log(y) - y/x]/[log(x) - x/y]
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derivative x^y=y^x
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To find the derivative y'(x), we can use implicit differentiation since x and y are both variables that depend on each other.
Step 1: Differentiate both sides of the equation with respect to x.
Using the chain rule, the derivative of the left side (x^y) with respect to x is:
d/dx (x^y) = d/dx (e^(ln(x^y))) [Applying the natural logarithm to rewrite x^y as e^(ln(x^y))]
= e^(ln(x^y)) * d/dx (ln(x^y)) [Using the chain rule]
= x^y * d/dx (ln(x^y)) [Since e^(ln(x^y)) = x^y]
Similarly, the derivative of the right side (y^x) with respect to x is:
d/dx (y^x) = y^x * d/dx (ln(y)) [Using the chain rule]
Step 2: Set the derivatives of both sides equal to each other.
Now, since x^y * d/dx (ln(x^y)) = y^x * d/dx (ln(y)), we can write:
x^y * (d/dx (ln(x^y))) = y^x * (d/dx (ln(y))) [Setting the derivatives equal to each other]
Step 3: Solve for y'(x).
To solve for y'(x), we need to express d/dx (ln(x^y)) and d/dx (ln(y)).
Recall that the derivative of ln(u) with respect to u is du/dx.
Therefore, d/dx (ln(x^y)) can be rewritten as:
d/dx (ln(x^y)) = (d/dx (x^y)) / (x^y) [Using the chain rule]
Similarly, d/dx (ln(y)) can be rewritten as:
d/dx (ln(y)) = (d/dx (y)) / y [Using the chain rule]
Substituting these expressions back into our equation, we have:
x^y * ((d/dx (x^y)) / (x^y)) = y^x * ((d/dx (y)) / y)
Simplifying this equation, we get:
(d/dx (x^y)) / x^y = (d/dx (y)) / y
Step 4: Solve for y'(x).
To solve for y'(x), we can multiply both sides of the equation by x^y * y, which eliminates the fractions:
(d/dx (x^y)) * y = (d/dx (y)) * x^y
Finally, dividing both sides of the equation by x^y gives us:
d/dx (x^y) = [(d/dx (y)) * x^y] / y
Therefore, y'(x) = [(d/dx (y)) * x^y] / y