z= x^2*y^3-y+1 find its tangent plane at point (1,1,1)
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$\displaystyle \frac{\partial z}{\partial x} = 2x$ $\displaystyle \frac{\partial z}{\partial y} = 3y^2 -1$ $\displaystyle f_{x}(1,1,1) = 2 $ $\displaystyle f_{y}(1,1,1) = 2$ $\displaystyle z-1= 2(x-1)+2(y-1) \Rightarrow z=2x+2y+3$
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