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Question Number 69766 by Rio Michael last updated on 27/Sep/19

$$find\frac{dy}{dx}atthepoint(0,3)when2x^{2}y+y+4{xy}^2=2x+3$$

Commented by kaivan.ahmadi last updated on 27/Sep/19

$$f(x,y)=2x^{2}y+y+4{xy}^2-2x-3=0$$$$\frac{dy}{dx}=-\frac{f_x^{\prime }}{f_y^{\prime }}=-\frac{4xy+4y^2-2}{2x^2+1+4x}{\mid }_{(0,3)}=-\frac{4{\left(3\right)}^2-2}{1}=-34$$

Commented by MJS last updated on 27/Sep/19

$$\mathrm{implicit}\mathrm{differentiation}$$$$\left(\frac{d}{dx}\left[f(x,y)\right]\right)dx=-\left(\frac{d}{dy}\left[f(x,y)\right]\right)dy$$$$\mathrm{on}\mathrm{the}\mathrm{l}.\mathrm{h}.\mathrm{s}\mathrm{treat}y\mathrm{as}\mathrm{a}\mathrm{constant}$$$$\mathrm{on}\mathrm{the}\mathrm{r}.\mathrm{h}.\mathrm{s}\mathrm{treat}x\mathrm{as}\mathrm{a}\mathrm{constant}$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{calculate}\frac{dx}{dy}\mathrm{and}\frac{dy}{dx}$$$$\mathrm{at}\mathrm{the}\mathrm{end}\mathrm{put}x=0,y=3$$$$\mathrm{in}\mathrm{this}\mathrm{case}$$$$f(x,y)=2x^{2}y+4{xy}^2+y-2x-3$$$$\mathrm{I}\mathrm{think}\mathrm{now}\mathrm{you}\mathrm{can}\mathrm{do}\mathrm{it}$$

Commented by Rio Michael last updated on 27/Sep/19

$$suresir$$$$4xy+2x^2\frac{dy}{dx}+4y^2+8xy\frac{dy}{dx}+\frac{dy}{dx}=2$$$$2x^2\frac{dy}{dx}+8xy\frac{dy}{dx}+\frac{dy}{dx}=2-4xy-4y^2$$$$\frac{dy}{dx}(2x^2+8xy+1)=2(1-2xy-2)$$$$\frac{dy}{dx}=\frac{2(1-2xy-2)}{2x^2+8xy+1}$$$$x=0andy=3$$$$\frac{dy}{dx}=\frac{2(1-2)}{1}=-2isthatcorrect?$$

Commented by MJS last updated on 27/Sep/19

$$\frac{d}{dx}[2x^{2}y+4{xy}^2+y-2x-3]=4xy+4y^2-2$$$$\frac{d}{dy}[2x^{2}y+4{xy}^2+y-2x-3]=2x^2+8xy+1$$$$(4xy+4y^2-2)dx=-(2x^2+8xy+1)dy$$$$\frac{dy}{dx}=-\frac{4xy+4y^2-2}{2x^2+8xy+1}$$$$x=0\wedge y=3\Rightarrow \frac{dy}{dx}=-\frac{34}{1}=-34$$

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