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Calc 2 help? 2 questions?

+3 votes
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.
asked Feb 22, 2013 in CALCULUS by payton Apprentice

2 Answers

+3 votes

f(x) = xe5x

Apply 'derivative with respect to x' each side.

f'(x) = (d/dx)(xe5x)

Derivative of Exponential Functions: (d/dx)(eax) = a eax

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

f'(x) = 1(e5x) + x(e5x)(5) = (e5x + 5xe5x) = (1 + 5x)(e5x) = 1·50 + 51(x)(e5x)

f''(x) = 5e5x + 5[(1(e5x) + 5x(e5x)] = 5e5x +5e5x + 25xe5x = (10 + 25x)e5x = (2·51 + 52x)5e5x

f'''(x) = 10(5e5x) + 25[e5x + 5xe5x] = 50e5x + 25e5x + 125e5x = 75e5x + 125e5x = (3·52 + ·53x)e5x

Continue this processes 'n' times then

fn(x) = [n·5n-1 + 5nx]e5x.

answered Feb 22, 2013 by britally Apprentice
0 votes

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- 2yx5 = (1/xy)[xy] '

- 2x6 e- 2y y ' + 6e- 2yx5 = (1/xy)[xy ' + y]

- 2x6 e- 2y y ' - (y '/y) = (1/x) - 6e- 2yx5

y ' [(- 2yx6 e- 2y y - 1)/y ] = (1- 6e- 2yx6)/x

y ' = [(1- 6e- 2yx6)y]/[(- 2yx6 e- 2y y - 1)x]

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

                  = (y/x)[(e2y- 6x6) / (- 2yx6y - e2y)].

                  =(y/x)[(6x6- e2y) / (2yx6y + e2y)].

answered Jul 29, 2014 by lilly Expert
edited Jul 29, 2014 by bradely

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