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