Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 74723 by necxxx last updated on 29/Nov/19

3xy+x^2 +y^2 =5 find the second derivative

$$\mathrm{3}{xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5} \\ $$$${find}\:{the}\:{second}\:{derivative} \\ $$

Commented by mathmax by abdo last updated on 29/Nov/19

y^2 +3xy+x^2 −5=0 Δ=(3x)^2 −4(x^2 −5) =9x^2 −4x^2 +20 =5x^2 +20 ⇒ y_1 (x)=((−3x+(√(5x^2 +20)))/2) and y_2 (x)=((−3x−(√(5x^2 +20)))/2) y=y_1 ⇒y^′ (x)=−(3/2)+(1/2)×((10x)/(2(√(5x^2 +20)))) =−(3/2) +(5/2)(x/(√(5x^2 +20))) and y^((2)) (x)=(5/2){x (5x^2 +20)^(−(1/2)) }^((1)) =(5/2){(5x^2 +20)^(−(1/2)) +x ×(−(1/2)(10x)(5x^2 +29)^(−(3/2)) } =(5/(2(√(5x^2 +20))))−((5x)/((5x^2 +20)(√(5x^2 +20)))) y=y_2 we follow the same method...

$${y}^{\mathrm{2}} +\mathrm{3}{xy}+{x}^{\mathrm{2}} −\mathrm{5}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{3}{x}\right)^{\mathrm{2}} −\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{5}\right)\:=\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{20}\:=\mathrm{5}{x}^{\mathrm{2}} \:+\mathrm{20}\:\Rightarrow \\ $$$${y}_{\mathrm{1}} \left({x}\right)=\frac{−\mathrm{3}{x}+\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}}}{\mathrm{2}}\:{and}\:{y}_{\mathrm{2}} \left({x}\right)=\frac{−\mathrm{3}{x}−\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}}}{\mathrm{2}} \\ $$$${y}={y}_{\mathrm{1}} \:\Rightarrow{y}^{'} \left({x}\right)=−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{10}{x}}{\mathrm{2}\sqrt{\mathrm{5}{x}^{\mathrm{2}} \:+\mathrm{20}}}\:=−\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{5}}{\mathrm{2}}\frac{{x}}{\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}}} \\ $$$${and}\:{y}^{\left(\mathrm{2}\right)} \left({x}\right)=\frac{\mathrm{5}}{\mathrm{2}}\left\{{x}\:\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\left\{\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:+{x}\:×\left(−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}{x}\right)\left(\mathrm{5}{x}^{\mathrm{2}} \:+\mathrm{29}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \right\}\right. \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}}}−\frac{\mathrm{5}{x}}{\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}\right)\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}}} \\ $$$${y}={y}_{\mathrm{2}} \:\:\:\:{we}\:{follow}\:{the}\:{same}\:{method}... \\ $$

Answered by Tanmay chaudhury last updated on 29/Nov/19

3x(dy/dx)+3y+2x+2y.(dy/dx)=0 [(dy/dx)=((−(3y+2x))/(3x+2y)) 3x×(d^2 y/dx^2 )+3(dy/dx)+3(dy/dx)+2+2y.(d^2 y/dx^2 )+2((dy/dx))^2 =0 pls put value of (dy/dx) and calculate

$$\mathrm{3}{x}\frac{{dy}}{{dx}}+\mathrm{3}{y}+\mathrm{2}{x}+\mathrm{2}{y}.\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\left[\frac{{dy}}{{dx}}=\frac{−\left(\mathrm{3}{y}+\mathrm{2}{x}\right)}{\mathrm{3}{x}+\mathrm{2}{y}}\right. \\ $$$$\mathrm{3}{x}×\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{3}\frac{{dy}}{{dx}}+\mathrm{3}\frac{{dy}}{{dx}}+\mathrm{2}+\mathrm{2}{y}.\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${pls}\:{put}\:{value}\:{of}\:\frac{{dy}}{{dx}}\:{and}\:{calculate} \\ $$

Commented by necxxx last updated on 29/Nov/19

Thank you so much sir.

$${Thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com