Given the equation y ln (x^2 + y^2 + 8) = 3​, evaluate dy/dx. Assume that the equation implicitly defines y as a differentiable function of

Question

Given the equation y ln (x^2 + y^2 + 8) = 3​, evaluate dy/dx. Assume that the equation implicitly defines y as a differentiable function of x.

in progress 0
Kylie 7 hours 2021-09-15T11:18:44+00:00 1 Answer 0

Answers ( )

    0
    2021-09-15T11:20:17+00:00

    y\ln(x^2+y^2+8)=3

    Differentiate both sides with respect to x:

    \dfrac{\mathrm dy}{\mathrm dx}\ln(x^2+y^2+8)+y\dfrac{\frac{\mathrm d}{\mathrm dx}[x^2+y^2+8]}{x^2+y^2+8}=0

    We have by the chain rule,

    \dfrac{\mathrm d}{\mathrm dx}[x^2+y^2+8]=2x+2y\dfrac{\mathrm dy}{\mathrm dx}

    so that

    *** QuickLaTeX cannot compile formula:
    \dfrac{\mathrm dy}{\mathrm dx}\left(\ln(x^2+y^2+8)+\dfrac{2y^2}{x^2+y^2+8}\right)=-\dfrac{2xy}{x^2+y^2+8}
    
    *** Error message:
    Error processing_dvigif_log(): -2
    

    \dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2xy}{(x^2+y^2+8)\left(\ln(x^2+y^2+8)+\frac{2y^2}{x^2+y^2+8}\right)}

    \dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2xy}{(x^2+y^2+8)\ln(x^2+y^2+8)+2y^2}

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )