Find tangent plane

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

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Tangent Plane to a Level Surface

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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4.4 Tangent Planes and Linear Approximations

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.

Using the gradient vector to find the tangent plane equation

The steps to populate the general equation of the tangent plane are as follows: Plug the values for x0 and y0 into the given function z = f ( x, y) to obtain the value for f ( x0, y0 ). Take the partial

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Tangent planes (article)

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