Jika y merupakan fungsi implisit dari x, carilah dy/dx dan d²y/dx².Diketahui, (x-y)pangkat² – y = 3 mohon bantuanya teman-teman, terima kasih





Jika y merupakan fungsi implisit dari x, carilah dy/dx dan d²y/dx².Diketahui, (x-y)pangkat² – y = 3 mohon bantuanya teman-teman, terima kasih





(x-y)^2-y=3\\\\
x^2-2xy+y^2-y=3\\\\
 \\frac{d}{dx}( x^2-2xy+y^2-y)= \\frac{d}{dx}( 3)\\\\
2x-(2x)\(y)+(2x)(y)\+2y.y\-1.y\=0\\\\
2x-(2y+2xy\)+2y.y\-y\=0\\\\
2x-2y-2xy\+2y.y\-y\=0\\\\
(-2x+2y-1)y\+2x-2y=0\\\\
(-2x+2y-1)y\=-2x+2y\\\\
 y\ = \\frac{-2x+2y}{-2x+2y-1} \\\\\\\\
y\= \\frac{2x-2y}{2x-2y+1} \\\\\\\\
 \\frac{dy}{dx} = \\frac{2x-2y}{2x-2y+1} \\\\\\\\


klo yang turunan kedua itu
 \\frac{d^2y}{dx^2}= \\frac{d}{dx}(y\)  \\\\\\\\
\\frac{d}{dx}(y\) =\\frac{d}{dx}(\\frac{2x-2y}{2x-2y+1} ) \\\\\\\\
 \\frac{d^2y}{dx^2} =\\frac{(2x-2y)\(2x-2y+1)-(2x-2y)(2x-2y+1)\}{(2x-2y+1)^2}

 \\frac{d^2y}{dx^2} =\\frac{(2-2.y\)(2x-2y+1)-(2x-2y)(2-2.y\)}{(2x-2y+1)^2}

 \\frac{d^2y}{dx^2} =\\frac{(4x-4y+2-4x.y\+4y.y\-2.y\)-(4x-4x.y\-4y+4y.y\)}{(2x-2y+1)^2}

 \\frac{d^2y}{dx^2} =\\frac{2-2.y\}{(2x-2y+1)^2}

 \\frac{d^2y}{dx^2} =\\frac{2-2(\\frac{2x-2y}{2x-2y+1})}{(2x-2y+1)^2}

 \\frac{d^2y}{dx^2} =\\frac{2-2(\\frac{2x-2y}{2x-2y+1})}{(2x-2y+1)^2}\\times  \\frac{(2x-2y+1)}{(2x-2y+1)} \\\\\\\\
 \\frac{d^2y}{dx^2} =\\frac{2(2x-2y+1)-2(2x-2y)}{(2x-2y+1)^3}

\\frac{d^2y}{dx^2} =\\frac{2(2x-2y+1)-2(2x-2y)}{(2x-2y+1)^3}\\\\\\\\
\\frac{d^2y}{dx^2} =\\frac{2}{(2x-2y+1)^3}



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