Implicit Differentiation question
x^y = y^x
I was wondering if I was right in the steps I was taking
first was to take the Ln or both sides making it
y Ln(x) = x Ln(y)
then just taking the derivatives of both sides using product rule twice
y/x + y'Ln(x) = xy'/y + ln(y)
then I would just solve for y'
am I doing any steps wrong?
I do not see anything wrong with the way you are going about it.
Alright thanks
Your steps are almost correct, but you made a mistake in the application of the product rule.
Here's the correct step-by-step process:
1. Start with the equation x^y = y^x.
2. Take the natural logarithm (ln) of both sides to simplify the equation:
ln(x^y) = ln(y^x).
3. Apply the logarithm property to bring down the exponents:
y ln(x) = x ln(y).
4. Differentiate both sides with respect to x using the implicit differentiation technique. Treat y as a function of x, so we'll need to use the chain rule:
(y ln(x))' = (x ln(y))'.
5. Apply the product rule on the left side:
y' ln(x) + y/x = (x ln(y))'.
6. Now, differentiate each term on the right side using the product and chain rules:
y' ln(x) + y/x = (x/y) * (y' / y) + ln(y) * (x/y)'.
7. Simplify the derivative on the right side:
y' ln(x) + y/x = (xy' - y) / y^2 + ln(y) * (-x/y^2) * y'.
8. Multiply through by yx to get rid of the denominators:
y'x ln(x) + y = xy' - y + ln(y) * (-x/y) * y'.
9. Rearrange the equation to solve for y':
y'x ln(x) - xy' + ln(y) * x = -y + y - ln(y) * x.
10. Simplify:
y'x ln(x) - xy' = - ln(y) * x.
11. Factor out y' on the left side:
y'(x ln(x) - x) = - ln(y) * x.
12. Finally, solve for y':
y' = (- ln(y) * x) / (x ln(x) - x).
So, your overall process was correct, but you made a mistake in applying the product rule on step 4. Make sure to double-check your differentiation steps to avoid such errors.