Calc 2 help? 2 questions?

Question

Suppose x^6 e−2y = ln (xy).
Find dy/dx by implicit differentiation.

Let f (x) = x e^5x.

Find a formula for the nth derivative of f, where n is any positive integer. Use x and n in your answer if needed.

Answer 1

f(x) = xe^5x

Apply 'derivative with respect to x' each side.

f'(x) = (d/dx)(xe^5x)

Derivative of Exponential Functions: (d/dx)(e^ax) = a e^ax

The Product Rule: (UV)' = U'V + UV'

f'(x) = 1(e^5x) + x(e^5x)(5) = (e^5x + 5xe^5x) = (1 + 5x)(e^5x) = 1·5⁰ + 5¹(x)(e^5x)

f''(x) = 5e^5x + 5[(1(e^5x) + 5x(e^5x)] = 5e^5x +5e^5x + 25xe^5x = (10 + 25x)e^5x = (2·5¹ + 5²x)5e^5x

f'''(x) = 10(5e^5x) + 25[e^5x + 5xe^5x] = 50e^5x + 25e^5x + 125e^5x = 75e^5x + 125e^5x = (3·5² + ·5³x)e^5x

Continue this processes 'n' times then

fⁿ(x) = [n·5^n-1 + 5ⁿx]e^5x.

Answer 2

The function is x⁶ e^{- 2y} = ln (xy).

Differentiate the function with respect to x.

The Product Rule: (UV)' = U'V + UV'

x⁶ (e^{- 2y}) ' + e^{- 2y}(x⁶) ' = [ln (xy)] '

x⁶ e^{- 2y} (- 2y ') + 6e^{- 2y}x⁵ = (1/xy)[xy] '

- 2x⁶ e^{- 2y} y ' + 6e^{- 2y}x⁵ = (1/xy)[xy ' + y]

- 2x⁶ e^{- 2y} y ' - (y '/y) = (1/x) - 6e^{- 2y}x⁵

y ' [(- 2yx⁶ e^{- 2y} y - 1)/y ] = (1- 6e^{- 2y}x⁶)/x

y ' = [(1- 6e^{- 2y}x⁶)y]/[(- 2yx⁶ e^{- 2y} y - 1)x]

dy/dx = y ' = (y/x)[(1- 6e^{- 2y}x⁶) / (- 2yx⁶ e^{- 2y} y - 1)].

                  = (y/x)[(e^2y- 6x⁶) / (- 2yx⁶y - e^2y)].

                  =(y/x)[(6x⁶- e^2y) / (2yx⁶y + e^2y)].