Limit Definition of Partial Derivatives (हिंदी में) | Calculus |

Welcome, friends, to this video inside, we have partial derivatives Limit definition description We will do some examples with it. As for us Know that f of derivative We will do it from x. Then y will be constant. In the definition, the value of y is not changing. Only x's value is changing And the value of x is changing from x plus h to x. and inside the denominator, the difference of x x plus h minus x It will be canceled out with x minus x. Then the denominator will be h. Similarly, we will take the partial derivative of f with respect to y. So the definition will be limit k approaches 0 (f of x, y plus k minus f of x, y) divided by k If f is partially differentiated with respect to y, then x is constant. In the definition, the value of x is not changing. Only y's value is changing That y plus k to y, then and in the denominator, the difference of y so y plus k minus y So y will cancel out with minus y. and the denominator will become k If partial derivative Any particular point Find a, b So in the definition, instead of x, y a, b will have to be replaced So the definition will be ∂f/∂x at (a, b) is equal to limit h approaches 0 f(a + h, b) minus f of a, b divided by h Similarly, del f by del y at a b is equal to the limit as k approaches 0 f(a, b + k) - f(a, b) divided by k Here, we will replace x with a and y with b in the definition. For example, using the definition of limit f of x, y equals We will find the partial derivative of \( x^2 - 4xy + 6 \) So, first partial data memorize the definition ∂f/∂x is equal to the limit as h approaches 0 f(x + h, y) minus f of x y divided by h So, to evaluate this limit, it should be required f(x + h, y) So instead of x We will replace x plus h in the definition of f Then we will get x plus h square minus 4y plus 6 minus f of x y x squared minus 4y plus 6 After opening the bracket, we will get x squared plus 2xh plus h squared minus 4y plus 6 The minus sign will go inside and become multiplication here. -x² + 4y - 6 So, plus x² minus x² will cancel out. will cancel out with -4y + 4y plus 6 minus 6 will cancel out And the remaining term will be 2x plus h squared, divided by h. h will cancel out from both the numerator and the denominator. And we'll have the limit as h approaches 0, 2x plus h, h equals 0, Then the value will be 2x. So the value of the partial derivative of f with respect to x will be 2x. Similarly, del f by del y, ∂f/∂y equals the limit as k approaches 0, (f of x, y plus k minus f of x, y) divided by k To evaluate this limit, we need to consider f of x, y, the value of y plus k. Inside the definition, we will replace y with y plus k. Then we will get x squared minus 4 y + k + 6 - x² - 4y + 6 After opening the bracket, x x² - 4y - 4k + 6 -x² + 4y - 6 So x squared minus x will cancel out. will cancel out with -4y + 4y plus 6 minus 6 will cancel out, and we will be left with minus 4k divided by k, k will cancel out with k So the final value of the limit will be minus 4. So ∂f/∂y equals minus 4 Example number 2, Using the limit definition of the partial derivative the function f of x y equals x y plus y squared 2, 3, take the partial derivative, first recall the definition, the partial derivative of f with respect to x at the particular point a, b limit h approaches 0 f of a plus h comma b minus f of a b divided by h Here point a b in the definition While we are given point 2 3 So, 2 3 will replace a b. Then it will come to us ∂f/∂x at (2, 3) is equal to limit h approaches 0 Replace a with 2 and b with 3 so f of 2 plus h 3 minus f of 2, 3 divided by h To evaluate this limit The value of the Cheyoge function 2 plus h And 3 on So we will replace x with the value 2 plus h And we will replace y with the value 3 inside the definition So we will get 2 plus h 3 plus 3 squared minus f of 2 3 We will substitute x with 2 and y with 3. so 2 into 3 plus 3 square After opening the bracket 6 plus 3 h plus 9 minus 6 -9 So, plus 6 minus 6 will cancel out. Plus 9 minus 9 will cancel out. So, it will come to us 3H divided by H It will be canceled by HH. So, the value we have is 3 So, the value of ∂f/∂x at 2, 3 is 3. ∂F/∂y We will evaluate on the 2nd and 3rd. So, in the definition We will place one at location 2 and two at location 3. So, ∂F by ∂Y 2,3 will be approached, limit k approaches 0 f of 2,3 plus k minus f of 2,3 Divide by k To evaluate this limit The value of the function is 2, 3 plus k If it goes to pay, then replace x with 2 and y's place 3 plus k If we replace it, our palas will come. 2.3 plus k plus 3 plus k whole square, minus x equals 2, y equals 3 so 2 into 3 plus 3 squared If we open the bracket, we will get 6 plus 2k plus 9 plus 6k plus k squared minus 6 minus 9 plus 6 minus 6 will cancel out, plus 9 minus 9 It will cancel out, and the value of 6 plus 2k will be 8k. So we will have 8k plus k squared divided by k. The k in the numerator will cancel out with the k in the denominator. Then we will have the limit k approach to 0.8 plus k, we will substitute the value of k 0, so 8 plus 0 equals 8, so ∂f/∂y The value arrived at 2 3 8, Example Number 3 limit of the partial derivative Using the definition Find the partial derivative of f(x, y) Where f of x y is 2x cubed plus x to the fourth power minus y squared divided by x squared plus y squared When x, y, 0, 0 is not the case If the value of x, y is 0, 0 So the value of the function is 0. And define partial at 0, 0 So definition ∂f/∂x at (0, 0) limit h approaches 0 f of 0 plus h comma 0 minus f of 0, 0 Then it will be limit as h tends to 0 f of h 0 minus f of 0, 0 divide by h To find the value of f of h 0 In place of x, h, and in place of y, We will replace 0 in the definition Then we will get 2h3 plus h4 minus 0 squared divide by s2 0 squared minus f of 00 The value 0 function If it is given, then it will remain with us. 2h3 plus h4 divide by s2, then divide by h Inside the numerator Here, s2 will cancel out. Then we will have it. 2 plus h divide by h Here, HH will be canceled. Then it will be limit H approaches 0, 2 plus H And the limit of H is 0. Then the value will be 2 plus 0 equals So the value at (0, 0) of del x by del x is 2. Similarly, del x by del y, 0, 0 at limit K tends to 0 F(0, 0) + K - F(0, 0) divided by K F of 0, K minus F of 0, 0 divided by K To evaluate this limit We will need the function's value to be 0k Replace x with 0 And k instead of y Then it will come to us. 00 minus k to the power of 5 divide by zero square plus k squared the value of minus f of 00 Given in the function 0 minus k to the power of 5 divided by k squared Power 5 divide by k squared The value will be minus k cubed Divide by k So limit as k approaches 0 minus k squared If the value of k is 0, then the value of the limit will become 0. Then ∂f/∂y at (0, 0) equals 0. Thank you for watching For more videos like this, share, and subscribe Raise the bell icon of the next-generation messenger