Differentiate y = (ln x)^ x + (sin x)^ x using logarithmic differentiation.

To differentiate y = ( x)^ x + ( x)^ x, we will use logarithmic differentiation for each term separately, then add the results. Let y = u + v, where u = ( x)^ x and v = ( x)^ x. Then (dy)/(dx) = (du)/(dx) + (dv)/(dx). Step 1: Differentiate u = ( x)^ x. Take the natural logarithm of both sides: u = (( x)^ x) Using the logarithm property (a^b) = b a: u = x · ( x) Now, differentiate both sides with respect to x using the product rule on the right side: (1)/(u) (du)/(dx) = (d)/(dx)( x) · ( x) + x · (d)/(dx)(( x)) We know (d)/(dx)( x) = - x. For (d)/(dx)(( x)), use the chain rule: (d)/(dx)( f(x)) = (f'(x))/(f(x)). Here f(x) = x, so f'(x) = (1)/(x). (d)/(dx)(( x)) = (1)/( x) · (1)/(x) = (1)/(x x) Substitute these derivatives back: (1)/(u) (du)/(dx) = - x · ( x) + x · (1)/(x x) Multiply by u to solve for (du)/(dx): (du)/(dx) = u (- x · ( x) + ( x)/(x x)) Substitute u = ( x)^ x back: (du)/(dx) = ( x)^ x (- x · ( x) + ( x)/(x x)) Step 2: Differentiate v = ( x)^ x. Take the natural logarithm of both sides: v = (( x)^ x) Using the logarithm property (a^b) = b a: v = x · ( x) Now, differentiate both sides with respect to x using the product rule on the right side: (1)/(v) (dv)/(dx) = (d)/(dx)( x) · ( x) + x · (d)/(dx)(( x)) We know (d)/(dx)( x) = (1)/(x). For (d)/(dx)(( x)), use the chain rule: (d)/(dx)(( x)) = (1)/( x) · (d)/(dx)( x) = (1)/( x) · x = x Substitute these derivatives back: (1)/(v) (dv)/(dx) = (1)/(x) · ( x) + x · x Multiply by v to solve for (dv)/(dx): (dv)/(dx) = v ((( x))/(x) + x · x) Substitute v = ( x)^ x back: (dv)/(dx) = ( x)^ x ((( x))/(x) + x · x) Step 3: Combine the derivatives. (dy)/(dx) = (du)/(dx) + (dv)/(dx) (dy)/(dx) = ( x)^ x (- x · ( x) + ( x)/(x x)) + ( x)^ x ((( x))/(x) + x · x) The final answer is ( x)^ x (- x · ( x) + ( x)/(x x)) + ( x)^ x ((( x))/(x) + x · x). Last free one today — make it count tomorrow, or type /upgrade for unlimited.