tangent\:of\:f(x)=\frac{1}{x^2},\:(-1,\:1) tangent\:of\:f(x)=x^3+2x,\:\:x=0; tangent\:of\:f(x)=4x^2-4x+1,\:\:x=1; tangent\:of\:y=e^{-x}\cdot \ln(x),\:(1,0) tangent\:of\:f(x)=\sin (3x),\:(\frac{\pi

Tangent Plane to a Surface. Let $(x_0 , y_0 , z_0 )$ be any point on the surface $z = f (x, y)$. If $f (x, y)$ is differentiable at $(x_0 , y_0 )$, then the surface has a tangent plane at $(x_0 , y_0 , z_0

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To find the tangent plane at the point, $P(x_o, y_o, z_o) = (-2, 1, 4)$, we’ll need to evaluate the partial derivatives of the functions at $(x_o, y_o)$. Evaluate the partial derivative with respect to $x$ by treating the variable, $y$, as a constant.

Find the equation of the tangent plane to the surface defined by the function f(x, y) = sin(2x)cos(3y) at the point (π/3, π/4). A tangent plane to a surface does not always exist at