1.3 KiB
Korzystając z różniczki wyznaczyć przybliżoną wartość wyrażenia.
1
\sqrt{2.01^{4}+9}=\left.5+df\right|_{(x_{0})}
x_0=2
\Delta x = 0.01
f(x)=\sqrt{x^4+9}
f'(x)=\frac{1}{2\sqrt{x^{4}+9}}\cdot4x^{3}
f'(x_{0})=3.2
\left.df\right|_{x_{0}}=3.2\cdot0.01=0.032
\sqrt{2.01^{4}+9}\approx5.032
2
(0.98^{2}+2)^3=27+\left.df\right|_{(x_{0})}
x_{0}=1
\Delta x=-0.02
f(x)=(x^{2}+2)^3
f'(x)=3(x^{2}+2)^{2}\cdot2x
f'(x_{0})=54
\left.df\right|_{x_{0}}=54\cdot(-0.02)=-1.08
(0.98^{2}+2)^{3}=27+(-1.08)=25.92
3
\cfrac{4\cdot2.03+1}{2.03^{2}-2}=\frac{9}{2}+\left.df\right|_{(x_{0})}
x_{0}=2
\Updelta x=0.03
f(x)=\cfrac{4x+1}{x^{2}-2}
f'(x)=\cfrac{1}{\left(x^{2}-2\right)^{2}}\cdot\left(4\cdot(x^{2}-2)-(4x+1)\cdot2x\right)
f'(x_{0})=-\cfrac{28}{4}=-7
\left.df\right|_{(x_{0})}=-7\cdot0.03=-0.21
\cfrac{4\cdot2.03+1}{2.03^{2}-2}=4.5+(-0.21)=4.29
4
\ln(1+\ln 1.01)=0+\left.df\right|_{x_{0}}
f(x)=\ln(1+\ln x)
x_{0}=1
\Delta x=0.01
f'(x)=\cfrac{1}{1+\ln x}\cdot\cfrac{1}{x}
f'(x_{0})=1
\left.df\right|_{x_{0}}=0.01
\ln(1+\ln 1.01)=0+0.01=0.01
5
0.99\cdot\cos(0.99\pi)=-1+\left.df\right|_{x_{0}}
f(x)=x\cdot\cos(x\pi)
x_0=1
\Delta x = -0.01
f'(x)=cos(x\pi)+x\cdot(-\sin(\pi x)\cdot\pi)
f'(x_{0})=\cos(\pi)-\pi(\sin\pi)=-1
\left.df\right|_{x_{0}}=-1\cdot(-0.01)=0.01
0.99\cdot\cos(0.99\pi)=-0.99