Question

НАЙТИ ПРОИЗВОДНЫЕ ФУНКЦИЙ В №50 (1,2)

Answer 1

f(x)=5In(x)+e^x

f⁾(x)=(5In(x)+e^x)⁾=(5/x) +e^x

f⁾(1)=(5/1) +e^1= 5+e

f⁾(x) =( (x³+1)/x)⁾ =( (x³+1)⁾*x-x⁾*(x³+1) )/x² = (3x³-x³-1)/x² =(2x³-1)/x²

f⁾(1) = (2*1³-1)/1² = 1

f(x)=3In(x)-x²

f⁾(x)=(3In(x)-x²)⁾= (3/x) -2x

f⁾(1)=(3/1)- 2*1= 1

f(x)=lgх +х³

f⁾(x)=(lgх +х³)⁾= 1/(х*ln(10)) +3x²

f⁾(-1)=1/( -1*ln(10))  +3*(-1)² = 3 - (1/ln10)

если считать ,что ln10≈2,3 то 3 - (1/ln10)= 3 - (1/2,3)=3 - 10/23= 59/23 =

= 2 13/23

Answer 2

Ответ:

$50.1)\ \ f(x)=3lnx-x^2\ \ ,\ \ \ f'(x)=\dfrac{3}{x}-2x\ \ ,\ \ f'(1)=3-2=1\\\\2)\ \ f(x)=lgx+x^3\ \ ,\ \ f'(x)=\dfrac{1}{x\, ln10}+3x^2\ \ ,\ \ f'(-1)=-\dfrac{1}{ln10}+3\\\\3)\ \ y=x^2\, lnx\ \ ,\ \ y'=2x\, lnx+x^2\cdot \dfrac{1}{x}=2x\, lnx+x=x\, (2\, lnx+1)\\\\4)\ \ y=(1-lnx)\, x\ \ ,\ \ y'=-\dfrac{1}{x}\cdot x+(1-lnx)\cdot 1=-1+1-lnx=-lnx\\\\5)\ \ f(z)=z^3-3\, lnz\ \ ,\ \ f'(z)=3z^2-\dfrac{3}{z}$

$6)\ \ f(x)=5\, lnx+e^{x}\ \ ,\ \ f'(x)=\dfrac{5}{x}+e^{x}\ \ ,\ \ f'(1)=5+e\\\\\\2)\ \ f(x)=\dfrac{x^3+1}{x}\ \ ,\ \ f'(x)=\dfrac{3x^2\cdot x-(x^3+1)}{x^2}=\dfrac{2x^3-1}{x^2}$