# 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
$0.99\cdot\cos(0.99\pi)=-1+\left.df\right|_{x_{0}}$
$f(x)=x\cdot\cos(x\pi)$
$x_0=1$
$\Delta x = -0.01$
$f'(x)=cos(x\pi)+x\cdot(-\sin(\pi x)\cdot\pi)$
$f'(x_{0})=\cos(\pi)-\pi(\sin\pi)=-1$
$\left.df\right|_{x_{0}}=-1\cdot(-0.01)=0.01$
$0.99\cdot\cos(0.99\pi)=-0.99$