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Question Number 188016 by Rupesh123 last updated on 24/Feb/23

Answered by aleks041103 last updated on 24/Feb/23

$$f={(x^3+y^2)}^{1/2}$$$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=$$$$=\frac{3x^2}{2\sqrt{x^3+y^2}}dx+\frac{y}{\sqrt{x^3+y^2}}dy=$$$$=\frac{1}{f}(\frac{3}{2}{x}^{2}dx+ydy)=df$$$$f(0.94,2.98)=$$$$=f(1,3)+f(0.94,2.98)-f(1,3)\approx f(1,3)+df$$$$df=\frac{1}{f(1,3)}(\frac{3\times 1^2\times (-0.06)}{2}+3\times (-0.02))=$$$$=\frac{1}{\sqrt{1^3+3^2}}(-0.09-0.06)=-\frac{0.15}{\sqrt{10}}$$$$\Rightarrow f(0.94,2.98)\approx \sqrt{10}-\frac{0.15}{\sqrt{10}}=0.985\sqrt{10}$$$$\Rightarrow f(0.94,2.98)\approx 0.985\sqrt{10}$$

Commented by Rupesh123 last updated on 24/Feb/23

df = # dx + # dy is form of the answer Your answer should be in the form. So please check your answer for the df. Your second answer is correct

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