# Find an equation of the tangent plane to the given surface at the specified point: z = e^(x^2-y^2), (1,-1,1)

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You need to remember the equation of the tangent plane to a given curve, at a given point `(x_0,y_0,z_0)` such that:

`z=f(x,y) = z_0 + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)`

You need to find the partial derivative `f_x(x,y),` hence, you need to differentiate the function `z=f(x,y)` with respect to x, considering y as constant, such that:

`f_x(x,y) = 2xe^(x^2-y^2)`

`f_x(1,-1) = 2e^(1-1) = 2e^0 = 2`

You need to find the partial derivative `f_y(x,y), ` hence, you need to differentiate the function `z=f(x,y) ` with respect to y, considering x as constant, such that:

`f_y(x,y) = -2ye^(x^2-y^2)`

`f_y(1,-1) = 2e^(1-1) = 2e^0 = 2`

**Hence, evaluating the equation of the tangent plane, under the given conditions yields `z = 2(x-1) + 2(y+1) + 1` .**