Step 1

If \(\displaystyle{y}={f{{\left({x}\right)}}}\) , then the independent variable x is called the input of the function and the dependent variable y is called the output of the function. The derivative of the function is the measure of how fast the function is changing with respect to x. A multi variable function \(\displaystyle{f{{\left({x},{y},{z}\right)}}}\) contains more than one inputs.

Step 2

The gradient of a multivariable function \(\displaystyle{f{{\left({x},{y}\right)}}}\) is defined as \(\displaystyle\nabla{f}-{\frac{{\partial{f}}}{{\partial{x}}}}{i}+{\frac{{\partial{f}}}{{\partial{y}}}}{j}\). To find gradient of \(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{y}\), substitute it in \(\nabla f-\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j\). According to standard results in partial differentiation \(\displaystyle{\frac{{\partial{x}^{{n}}}}{{\partial{x}}}}={n}{x}^{{{n}-{1}}}\) and \(\displaystyle{\frac{{\partial{y}^{{n}}}}{{\partial{x}}}}={n}{y}^{{{n}-{1}}}\), where n is an integer.

\(\displaystyle\nabla{f}={\frac{{\partial{f}}}{{\partial{x}}}}{i}+{\frac{{\partial{f}}}{{\partial{y}}}}{j}\)

\(\displaystyle={i}{\frac{{\partial}}{{\partial{x}}}}{\left({x}^{{2}}-{y}\right)}+{j}{\frac{{\partial}}{{\partial{y}}}}{\left({x}^{{2}}-{y}\right)}\)

\(\displaystyle={2}\xi-{j}\)

To find the gradient at a particular point \(\displaystyle{\left({x}_{{0}},{y}_{{0}}\right)}\), substitute it in \(\displaystyle\nabla{f}={2}\xi-{j}\) and find the corresponding vectors. Thus the value of gradient corresponding to the points that lies on the y axis can be calculated by substituting \(\displaystyle{y}={0}\) in \(\displaystyle\nabla{f}={2}\xi-{j}\)

\(\displaystyle\nabla{f}={2}\xi-{j}\)

\(\displaystyle={0}{i}-{j}\)

\(\displaystyle=-{j}\)