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Question Number 190249 by alcohol last updated on 30/Mar/23

$$if8\mathrm{\%}errorismadeonx,$$$$whatisthepercentageerror$$$$on\lambda x^{\frac{1}{5}}?$$

Answered by mr W last updated on 30/Mar/23

$$y=\lambda x^{\frac{1}{5}}$$$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\approx \frac{dy}{dx}=\frac{\lambda }{5}{x}^{-\frac{4}{5}}$$$$\frac{\mathrm{\Delta }y}{y}=\frac{\frac{\lambda }{5}{x}^{-\frac{4}{5}}}{\lambda x^{\frac{1}{5}}}=\frac{\mathrm{\Delta }x}{5x}=\frac{8\mathrm{\%}}{5}=1.6\mathrm{\%}$$$$i.e.erroron\lambda x^{\frac{1}{5}}is1.6\mathrm{\%}.$$

Commented by mr W last updated on 30/Mar/23

$$or$$$$y=\lambda x^{\frac{1}{5}}$$$$y+\mathrm{\Delta }y=\lambda {(x+\mathrm{\Delta }x)}^{\frac{1}{5}}\approx \lambda x^{\frac{1}{5}}(1+\frac{\mathrm{\Delta }x}{5x})=y(1+\frac{\mathrm{\Delta }x}{5x})$$$$\mathrm{\Delta }y\approx y\frac{\mathrm{\Delta }x}{5x}$$$$\frac{\mathrm{\Delta }y}{y}\approx \frac{\mathrm{\Delta }x}{5x}=\frac{8\mathrm{\%}}{5}=1.6\mathrm{\%}$$

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