"Since \\ h(x,y)=yln(x), we \\ get \\\\\nh_x(x,y)=\\frac{y}{x} \\ and \\\\\nh_y(x,y)=\\ln(x).\\\\\nat (1,5) , we \\ obtain\\\\\n\nh_x(1,5)= 5\\ and \\\\\nh_y(1,5)=\\ln(1)=0.\\\\\nAs \\ at (1,5)\n \\text{the equation of }\\\\ \\text{the tangent plane to }h(x,y) is \\\\\nz=h_x (1,5)(x-1)+h_y(1,5)(y-5)\\\\\nSo, we \\ get \\\\\nz=5(x-1).\\\\\n\\text{This implies that the correct answer is }\\\\\nf."
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