Derivatives Question?

Question

I am confused on how to find the first derivatives of this equation, I already know the answer, but would like help on how to get there.

f(x)=x^3 e^(-0.01x)

Answer 1

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f(x) = (x^3)e^(-0.01x)

The function is of the form f(x) = u(x)v(x) and the derivative is of the form

f'(x) = u'v + uv'

If we take u = x^3 and v = e^(-0.01x) we have

u' = 3x^2

v' = -0.01e^(-0.01x)

f'(x) = (3x^2)e^(-0.01x) -0.01(x^3)e^(-0.01x)

 = (3- 0.01x)(x^2)e^(-0.01x)

Answer 2

:-

f ( x ) = x³ e^(- 0•01x)

f ' (x) = 3x² e^(- 0•01x) - (0•01) x³ e^(-0•01 x)

f ' (x) = x² e^(- 0•01x) [ 3 - 0•01 x ]

Answer 3

f(x)=x^3 e^(-0.01x). Take log on both sides

log f(x)= log x^3 -0.01x= 3 log x -0.01x. Now differentiate

1/f df/dx= 3/x -0.01

df/dx= (3/x-0.01) f(x)= (3/x-0.01) (x^3 e^(-0.01x))

This is the answer.

Answer 4

f'(x) = x^3*d/dx [e^(-0.01x)] + e^(-0.01x)*d/dx(x^3)

= -0.01 x^3*e^(-0.01x) + 3x^2 * e^(-0.01x)

= (3x^2 - 0.01x^3) * e^(-0.01x).