The derivative of $$f(x, y)$$ at point $$(1, 2)$$ in the direction of vector $$\overrightarrow i + \overrightarrow j $$ is $$2\sqrt 2 $$ and in the direction of the vector $$ - 2\overrightarrow j $$ is $$-3.$$ Then the derivative of $$f(x,y)$$ in direction $$ - \overrightarrow i - 2\overrightarrow j $$ is

The differential equation $${y^{11}} + {\left( {{x^3}\,\sin x} \right)^5}{y^1} + y = \cos {x^3}\,\,\,\,$$ is

The solution of a differential equation $${y^{11}} + 3{y^1} + 2y = 0$$ is of the form

Let $$\,\,f\left( x \right) = x - \cos \,x.\,\,\,$$ Using Newton-Raphson method at the $$\,{\left( {n + 1} \right)^{th}}$$ iteration, the point $$\,{x_{n + 1}}$$ is computed from $${x_n}$$ as