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calculus: if f(x)= ln(x+4+e^-3x), then f ' (0) is?

please explain/show all work

thank you

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The Best Answer for calculus: if f(x)= ln(x+4+e^-3x), then f ' (0) is?

f(x)= ln(x+4+e^-3x)

remembering d(ln(u))/du = (1/u)*du

f'(x) = 1/(x+4+e^(-3x))*(1-3e^(-3x)) = (1-3e^(-3x))/(x+4+e^(-3x))

f'(0) = (1-3e^(-3*0))/(0 + 4 + e^(-3*0))

f'(0) = (1-3*1)/(4+1)

f'(0) = -2/5 --> Final Answer

Happy mathing!

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f'(x) = (1−3e^−(3x))/(e^−(3x)+x+4)

is the derivative of that function

it's chain rule, get the derivative of ln(x+4+e^-3x) first, which is just 1/(x+4+e^-3x) then multiply it by the derivative of x + 4 + e^-3x which is just 1 - 3e^-3x

plug 0 in for all the x values and your solution is -2/5

I think this is the correct answer

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f(x) = ln(x + 4 + e^-3x)

df/dx = [1 + 0 - 3e^-3x]/(x + 4 + e^-3x) [= g'/g] where g = (x+4+e^-3x)

df/dx when x = 0 = [1 - 3]/5

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