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Question Number 31442 by mondodotto@gmail.com last updated on 08/Mar/18
Commented by mondodotto@gmail.com last updated on 08/Mar/18
differentiate
$$\boldsymbol{\mathrm{differentiate}} \\ $$
Commented by prof Abdo imad last updated on 09/Mar/18
we have −2y=x^(−2) (x^2 +4)^(1/2) ⇒ −2y^′ = −2x^(−3) (x^2 +4)^(1/2) +x^(−2) x(x^2 +4)^(−(1/2)) −2y^′ = ((−2(√(x^2 +4)))/x^3 ) + (1/(x(√(x^2 +4)))) ⇒ y^′ (x)= ((√(x^2 +4))/x^3 ) −(1/(2x(√(x^2 +4)))) = ((2x(x^2 +4)−x^3 )/(2x^4 (√(x^2 +4)))) = ((x^3 +8x)/(2x^4 (√(x^2 +4)))) y^′ (x)= ((x^2 +8)/(2x^3 (√(x^2 +4)))) .
$${we}\:{have}\:\:−\mathrm{2}{y}={x}^{−\mathrm{2}} \:\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow \\ $$$$−\mathrm{2}{y}^{'} =\:−\mathrm{2}{x}^{−\mathrm{3}} \left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+{x}^{−\mathrm{2}} \:{x}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$−\mathrm{2}{y}^{'} =\:\frac{−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{{x}^{\mathrm{3}} }\:\:+\:\:\frac{\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)=\:\frac{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}{{x}^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}} \\ $$$$=\:\:\frac{\mathrm{2}{x}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)−{x}^{\mathrm{3}} }{\mathrm{2}{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$=\:\:\frac{{x}^{\mathrm{3}} \:+\mathrm{8}{x}}{\mathrm{2}{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}} \\ $$$${y}^{'} \left({x}\right)=\:\frac{{x}^{\mathrm{2}} \:+\mathrm{8}}{\mathrm{2}{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}\:. \\ $$
Answered by MJS last updated on 08/Mar/18
f(x)=((g(x))/(h(x))) ⇒ f′(x)=((g′(x)×h(x)−g(x)×h′(x))/(h^2 (x))) g(x)=−(√(x^2 +4))=−(x^2 +4)^(1/2) h(x)=2x^2 g(x)=i(j(x)) ⇒ g′(x)=i′(j(x))×j′(x) i(x)=−x^(1/2) ⇒ i′(x)=−(1/2)x^(−(1/2)) j(x)=x^2 +4 ⇒ j′(x)=2x g′(x)=−(1/2)(x^2 +4)^(−(1/2)) ×2x=−x×(x^2 +4)^(−(1/2)) h′(x)=4x g′(x)×h(x)=−2x^3 (x^2 +4)^(−(1/2)) −g(x)×h′(x)=4x×(x^2 +4)^(1/2) y′=((−2x^3 (x^2 +4)^(−(1/2)) +4x×(x^2 +4)^(1/2) )/(4x^4 ))= =((−x^2 (x^2 +4)^(−(1/2)) +2(x^2 +4)^(1/2) )/(2x^3 ))= =(((x^2 +4)^(−(1/2)) (−x^2 +2(x^2 +4)))/(2x^3 ))= =((x^2 +8)/(2x^3 (√(x^2 +4))))
$${f}\left({x}\right)=\frac{{g}\left({x}\right)}{{h}\left({x}\right)}\:\Rightarrow\:{f}'\left({x}\right)=\frac{{g}'\left({x}\right)×{h}\left({x}\right)−{g}\left({x}\right)×{h}'\left({x}\right)}{{h}^{\mathrm{2}} \left({x}\right)} \\ $$$${g}\left({x}\right)=−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}=−\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${h}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} \\ $$$${g}\left({x}\right)={i}\left({j}\left({x}\right)\right)\:\Rightarrow\:{g}'\left({x}\right)={i}'\left({j}\left({x}\right)\right)×{j}'\left({x}\right) \\ $$$${i}\left({x}\right)=−{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow\:{i}'\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${j}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}\:\Rightarrow\:{j}'\left({x}\right)=\mathrm{2}{x} \\ $$$${g}'\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{2}{x}=−{x}×\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${h}'\left({x}\right)=\mathrm{4}{x} \\ $$$${g}'\left({x}\right)×{h}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$−{g}\left({x}\right)×{h}'\left({x}\right)=\mathrm{4}{x}×\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${y}'=\frac{−\mathrm{2}{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{4}{x}×\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{4}{x}^{\mathrm{4}} }= \\ $$$$=\frac{−{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}{x}^{\mathrm{3}} }= \\ $$$$=\frac{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(−{x}^{\mathrm{2}} +\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}\right)\right)}{\mathrm{2}{x}^{\mathrm{3}} }= \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{2}{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$ \\ $$

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