Answer:
We have, \[\sqrt{25.2}\] Consider\[f\,(x)=\sqrt{x}\] Let \[x=25\]and \[\Delta \,x=0.2\] \[\therefore \] \[f'(x)=\frac{1}{2\sqrt{x}}\] Now, \[f\,(x+\Delta \,x)\approx f\,(x)+\Delta \,xf'(x)\] \[\Rightarrow \] \[\sqrt{x+\Delta \,x}\approx \sqrt{x}++\frac{1}{2\sqrt{x}}\cdot \Delta \,x\] \[\Rightarrow \] \[\sqrt{25.2}\approx \sqrt{25}+\frac{0.2}{2\sqrt{25}}=5+\frac{0.2}{10}=5.02\] \[\Rightarrow \] \[\sqrt{25.2}=5.02\]
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