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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

$$\mathbf{differentiate}$$

Commented by prof Abdo imad last updated on 09/Mar/18

$$wehave-2y=x^{-2}{(x^2+4)}^{\frac{1}{2}}\Rightarrow$$$$-2y^{{}^{\prime }}=-2x^{-3}{(x^2+4)}^{\frac{1}{2}}+x^{-2}x{(x^2+4)}^{-\frac{1}{2}}$$$$-2y^{{}^{\prime }}=\frac{-2\sqrt{x^2+4}}{x^3}+\frac{1}{x\sqrt{x^2+4}}\Rightarrow$$$$y^{{}^{\prime }}\left(x\right)=\frac{\sqrt{x^2+4}}{x^3}-\frac{1}{2x\sqrt{x^2+4}}$$$$=\frac{2x(x^2+4)-x^3}{2x^4\sqrt{x^2+4}}$$$$=\frac{x^3+8x}{2x^4\sqrt{x^2+4}}$$$$y^{{}^{\prime }}\left(x\right)=\frac{x^2+8}{2x^3\sqrt{x^2+4}}.$$

Answered by MJS last updated on 08/Mar/18

$$f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}\Rightarrow f^{\prime }\left(x\right)=\frac{g^{\prime }\left(x\right)\times h\left(x\right)-g\left(x\right)\times h^{\prime }\left(x\right)}{h^2\left(x\right)}$$$$g\left(x\right)=-\sqrt{x^2+4}=-{(x^2+4)}^{\frac{1}{2}}$$$$h\left(x\right)=2x^2$$$$g\left(x\right)=i\left(j\left(x\right)\right)\Rightarrow g^{\prime }\left(x\right)=i^{\prime }\left(j\left(x\right)\right)\times j^{\prime }\left(x\right)$$$$i\left(x\right)=-x^{\frac{1}{2}}\Rightarrow i^{\prime }\left(x\right)=-\frac{1}{2}{x}^{-\frac{1}{2}}$$$$j\left(x\right)=x^2+4\Rightarrow j^{\prime }\left(x\right)=2x$$$$g^{\prime }\left(x\right)=-\frac{1}{2}{(x^2+4)}^{-\frac{1}{2}}\times 2x=-x\times {(x^2+4)}^{-\frac{1}{2}}$$$$h^{\prime }\left(x\right)=4x$$$$g^{\prime }\left(x\right)\times h\left(x\right)=-2x^3{(x^2+4)}^{-\frac{1}{2}}$$$$-g\left(x\right)\times h^{\prime }\left(x\right)=4x\times {(x^2+4)}^{\frac{1}{2}}$$$$y^{\prime }=\frac{-2x^3{(x^2+4)}^{-\frac{1}{2}}+4x\times {(x^2+4)}^{\frac{1}{2}}}{4x^4}=$$$$=\frac{-x^2{(x^2+4)}^{-\frac{1}{2}}+2{(x^2+4)}^{\frac{1}{2}}}{2x^3}=$$$$=\frac{{(x^2+4)}^{-\frac{1}{2}}(-x^2+2(x^2+4))}{2x^3}=$$$$=\frac{x^2+8}{2x^3\sqrt{x^2+4}}$$$$$$

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