An equation of the tangent plane to the surface \(z=f(x,y)\) at the point \((x_{0},y_{0},z_{0})\) is given by
$$z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$$
Notice that the tangent plane to the surface \(z=f(x,y)\) at the point \((x_{0},y_{0},z_{0})\) is defined to be the plane that contains both tangent lines \(f_{x}(x_{0},y_{0})\) and \(f_{y}(x_{0},y_{0})\).
Example
Find an equation of the tangent plane to \(z=3x^{2}+5y^{2}+7\) at the point \((1,1,15)\).
Solution: We have
$$ f_{x}(x,y)=6x\ \ \ \ \ \text{and}\ \ \ \ \ f_{x}(1,1)=6$$
$$ f_{y}(x,y)=10y\ \ \ \ \ \text{and}\ \ \ \ \ f_{y}(1,1)=10$$
Therefore, the equation of the tangent plane at \((1,1,15)\) is
$$z-15=6(x-1)+10(y-1)\ \ \ \ \ \text{or}\ \ \ \ \ z=6x+10y-1$$
Proof
Suppose that a surface S has \(z=f(x,y)\), where f has continuous first derivatives, and let P\((x_{0},y_{0},z_{0})\) be the point on S.
Then we first recall that any plane passing through the point \((x_{0},y_{0},z_{0})\) has an equation of the form
$$A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0\ \ \ \ \ \cdots\ \ \text{(1)}$$
By dividing this equation by \(C\) and letting \(a=-\frac{A}{C}\) and \(b=-\frac{B}{C}\), we can rewrite (1) in the form
$$z-z_{0}=a(x-x_{0})+b(y-y_{0})\ \ \ \ \ \cdots\ \ \text{(2)}$$
Then, since the equation (2) represents the tangent plane at the point \((x_{0},y_{0},z_{0})\), its intersection with \(y=y_{0}\) is tangent line of \(f(x,y_{0})\) with slope \(f_{x}(x_{0}, y_{0})\).
Setting \(y=y_{0}\) in the equation (2), we have
$$z-z_{0}=a(x-x_{0})\ \ \ \text{if}\ \ \ y=y_{0}$$
and we see that this is a line equation with slope \(a\). Thus, we get \(a=f_{x}(x_{0},y_{0})\).
Similarly, setting \(x=x_{0}\) in the equation (2), we get \(z-z_{0}=b(y-y_{0})\) which represents the tangent line of \(f(x_{0},y)\) with slope \(f_{y}(x_{0}, y_{0})\). Thus, \(b=f_{y}(x_{0},y_{0})\).
Therefore, we get
$$z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$$