Answer to Question #118267 in Calculus for Olivia

Question #118267
The equation of the tangent plane to h(x,y)=yln(x) at the point (1,5) is
Select one:
a. z=1/5x+1/5y

b. z=x+y


c. z=1/4x+1/4y


d. z=4(x−1)


e. z=5(x+1)


f. z=5(x−1)
1
Expert's answer
2020-05-28T18:00:22-0400

"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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