Dear student Since you have not mentioned the question number so doing as per my understanding Consider the function y=fx=xLet x=25 and x+△x=25 3 Then,△x=25 3-25=0 3For x=25, we have y=25=5 Putting x=25 in y=xLet dx=△x=0 3Now, y=x⇒dydx=12x⇒dydxx=25=12×5=110∴dy=dydxdx⇒dy=1100 3=0 03⇒△y=0 03 ∵△y≅dyHence, 25 3=y+△y=5+0 03=5 03 For remaining queries we request you to post them in separate threads to have rapid assistance from our experts Regards

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Solve this: diff erentials, find the approximate value of 25.3 point on the curve Y2 = 4x which is nearest to the poil I the -8) the shortest distance between the two given lines : and 3i-2j-f+ß(3i+2j-4i) the equation of plane passing through the points (O, —l, —1' 1) and (3, 9, 4) graphically (Linear programming problem) lize Z = — 3x + 4y subJect to constraints, x+2yS8, 3x+2ys 12, x20,y20 e following Linear programming problem graphically Ee Z = 6x + 5y subject to constraints, 3x+5y S 15, 5x+2y S 10,

Dear student
Since you have not mentioned the question number so doing as per my understanding.

Consider the function y = f (x) = \sqrt{x} Let x = 25 and x + △ x = 25.3 Then, △ x = 25.3 - 25 = 0.3 For x = 25, we have y = \sqrt{25} = 5 [Putting x = 25 in y = \sqrt{x}] Let dx = △ x = 0.3 Now, y = \sqrt{x} \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{dy}{dx})}_{x = 25} = \frac{1}{2 \times 5} = \frac{1}{10} ∴ dy = \frac{dy}{dx} dx \Rightarrow dy = \frac{1}{10} (0.3) = 0.03 \Rightarrow △ y = 0.03 [∵ △ y ≅ dy] Hence, \sqrt{25.3} = y + △ y = 5 + 0.03 = 5.03

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