# find derivatives of the following function using differentiation rules HELP! f(x)= x^4 + 3x^2+4x-31 f(x)=2(x+3)e^x f(x)= (2x-1)e^x-3

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To find the derivatives of the following:

i)

f(x)= x^4 + 3x^2+4x-31

Differentiating both sides, we get:

f'(x) = (x^4+3x^2+4x-31)'

f'(x) = (x^4)' +(3x^2)'+(4x)'- (31)', as (f(x)+g(x))' = f'(x)+g'(x).

f'(x) = 4x^3+3*2x+4-0, as d/dx(k x^n ) = knx^(n-1).

f'(x) = 4x^3+6x+4.

ii)

f(x)=2(x+3)e^x.

Differentiating both sides, we get:

f'(x) ={2(x+3)e^x}'

f'(x) = {2(x+3)}' e^x +(2(x+3)(e^x)'

f'(x) = 2e^x+2(x+1)e^x , as (e^x)' = e^x.

f'(x) = 2(x+4)e^x.

iii0

f(x)= (2x-1)e^x-3 . We assume -3 is separate not in power of e.)

Differentiating both sides, we get:

f'(x) = (2x-1)'e^x+(2x-1)(e^x)' -(3)'

f'(x) = 2e^x +(2x-1)e^x-0

f'(x) = 2e^x+2xe^x -e^x.

f'(x) =(2x+1)e^x.