"We \\ have\\ h(x,y)= y ln(x)\\\\\n\\text{So, we get }\\\\\nh_x(x,y)=\\frac{y}{x}\\\\\n\u200b\t\n and\\\\\nh_y (x,y)=ln(x).\\\\\nat\\ (1,5), \\ we \\ get\\\\\nh _x (1,5)=5 \\ and\\\\\nh _y (1,5)=ln(1)=0.\\\\\nAs \\ at(1,5)\\ the \\ equation \\ of \\\\\n\\text{the tangent plane to } h(x,y)\\ is\\\\\nz=h_x (1,5)(x\u22121)+h_y (1,5)(y\u22125)\\\\\n\\text{So, we get} \\\\\nz=5(x\u22121).\\\\\n\\text{This implies that the correct answer is}\\\\\nf."
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