Consider the function
z(x,y)=e3x−6+xy+ln(y2).
We determine the tangent line through (x,y)=(2,1).
z′x(x,y)z′y(x,y)=3e3x−6+yx+2y.
slope=−3e3⋅2−6+12+21=−44=−1.
The general form of a tangent line is t(x)=ax+b. Now it holds that a=−1.
In order to determine b se use that the tangent line goes through (x,y)=(2,1). Hence, t(2)=−1⋅2+b=1, which results in b=3.
Hence, the tangent line is given by t(x)=−x+3.
z(x,y)=e3x−6+xy+ln(y2).
We determine the tangent line through (x,y)=(2,1).
z′x(x,y)z′y(x,y)=3e3x−6+yx+2y.
slope=−3e3⋅2−6+12+21=−44=−1.
The general form of a tangent line is t(x)=ax+b. Now it holds that a=−1.
In order to determine b se use that the tangent line goes through (x,y)=(2,1). Hence, t(2)=−1⋅2+b=1, which results in b=3.
Hence, the tangent line is given by t(x)=−x+3.