We can simply find the solution as follows, |

The function is well defined at (0,0) (2) check for the second step, That means the limit exists at (0,0) Now check step-3:
So that the function is continuous at origin. |

2. Here f1 = f2 3. now put y = mx, we get Here f1 = f2 = f3 Now put y = mx² 4. Therefore , F1 = f2 = f3 =f4 We can say that the limit exists with 0. |

To calculate treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules, [sin(y²x + 5x – 8)] = cos(y²x + 5x – 8)(y²x + 5x – 8) = (y² + 50)cos(y²x + 5x – 8) Similarly partial derivative of f(x,y) with respect to y is, [sin(y²x + 5x – 8)] = cos(y²x + 5x – 8)(y²x + 5x – 8) = 2xycos(y²x + 5x – 8) |

if , then show that- |

Here we have, u = …………………..(1) now partially differentiate eq.(1) w.r to x and y , we get = Or ………………..(2) And now,
= ………………….(3) Adding eq. (1) and (3) , we get = 0 Hence proved. |

Partially differentiating given equation with respect to and x and y then equate them to zero On solving above we get Also Thus we get the pair of values (0,0), (,0) and (0, Now, we calculate At the point (0,0) So function has saddle point at (0,0). At the point ( So the function has maxima at this point (. At the point (0, So the function has minima at this point (0,. At the point ( So the function has an saddle point at ( |

Let be the point on sphere which is nearest to the point . Then shortest distance. Let Under the condition … (1) By method of Lagrange’s undetermined multipliers we have … (2) … (3) i.e. & … (4) From (2) we get From (3) we get From (4) we get Equation (1) becomes i.e. y = 2 |

Suppose and Now taking L.H.S, Which is Hence proved.
2. So that |

If then prove that grad u , grad v and grad w are coplanar. |

Here-
Now- Apply
Which becomes zero. So that we can say that grad u, grad v and grad w are coplanar vectors. |

Here Now, And We know that- So that- Now, Directional derivative = |