If y = (x + √x2 + 1)2, show that dydx = 2y√x2 + 1

Sure, let's solve this step by step:

Step 1: Given y = (x + √(x² + 1))²

Step 2: Differentiate y with respect to x using the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

dy/dx = 2(x + √(x² + 1)) * d/dx (x + √(x² + 1))

Step 3: Now, differentiate the inner function (x + √(x² + 1)) with respect to x. The derivative of x is 1 and the derivative of √(x² + 1) is (1/2√(x² + 1)) * 2x (using the chain rule again).

So, d/dx (x + √(x² + 1)) = 1 + x/√(x² + 1)

Step 4: Substitute this back into the equation from Step 2:

dy/dx = 2(x + √(x² + 1)) * (1 + x/√(x² + 1))

Step 5: Simplify the equation:

dy/dx = 2(x + √(x² + 1)) + 2x(x + √(x² + 1))/√(x² + 1)

Step 6: Notice that (x + √(x² + 1)) is y, so substitute y back into the equation:

dy/dx = 2y + 2xy/√(x² + 1)

Step 7: Factor out 2y:

dy/dx = 2y(1 + x/√(x² + 1))

Step 8: Simplify the equation to get the final answer:

dy/dx = 2y√(x² + 1)