Question

Sa se calculeze derivata functiei f si apoi f'(x0) , cu x0 indicat :

a) f(x) = 3x⁵-2x³+4x+6, x indice 0 = 0

b) f(x) = 1+2x+3x²+....+nx ^ n-1 , n apartine N star

c) f(x) = xlnx , x indice 0 = e ​

Va rog dau coroana urgent

Answer 1

Explicație pas cu pas:

a) f(x) = 3x⁵ - 2x³ + 4x + 6

f'(x) = (3x⁵-2x³+4x+6)' =

= (3x⁵)' - (2x³)' + (4x)' + (6)'

= 3(x⁵)' - 2(x³)' + 4(x)' + 0

= 3•5x⁴ - 2•3x² + 4•1

= 15x - 6x + 4

f'(0) = 15•0 - 6•0 + 4 = 0 - 0 + 4 = 4

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b) f(x) = 1 + 2x + 3x² +....+ nxⁿ⁻¹

f'(x) = (1 + 2x + 3x² + 4x³ +....+ nxⁿ⁻¹)'

= 2 + 6x + 12x² + ... + n(n-1)xⁿ⁻²

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c) f(x) = x•ln(x)

f'(x) = (x•ln(x))' = x'•ln(x) + (ln(x))'•x

$= 1 \cdot ln(x) + \frac{1}{x} \cdot x = ln(x) + 1$

=> f'(x) = ln(x) + 1

f'(e) = ln(e) + 1 = 1 + 1 = 2

Answer 2

Răspuns:

a)f(x)=3x⁵-2x³+4x+6

f `(x)=3*5x⁴-2*3x²+4=

15x⁴-6x²+4

f(0)=15*0-6*0+4=0+4=4

b)1+2x+3x²+...+nxⁿ⁻¹

f `(x)=1 `+2x `+3x² `+...+nxⁿ⁻1  `=

0+2+3*2x+....+n(n-1)xⁿ-²=

2+6x+...+n(n-1)xⁿ⁻²

f `(0)=2+6*0+...+n(n-1)*0ⁿ⁻²=

2+0+...+0=2

c)f(x)=xlnx

f `(x)=lnx+x*1/x=lnx+1

f `(e)=lne+1=1+1=2

Explicație pas cu pas: