Find the Differential of a Function: y = √ (1 + x ^ 2) arctan x

Consider the function y = √ (1 + x²) * arctgx. At the request of the task, we find the differential of this function. As is known, if a differentiable function y = f (x) is given, then its differential dy is determined by the formula dy = f Ꞌ (x) dx. We calculate the derivative yꞋ = (√ (1 + x²) * arctgx) Ꞌ.
Let’s use the formula (u * v) ꞌ = uꞌ * v + u * vꞌ. Then, we get: yꞋ = (√ (1 + x²)) Ꞌ * arctgx + √ (1 + x²) * (arctgx) Ꞌ = ((1 + x²) ½) Ꞌ * arctgx + √ (1 + x²) * ( arctgx) Ꞌ. We use the corresponding differentiation properties and the following formulas (un) ꞌ = n * un – 1 * uꞌ, where n is a constant and (arctgx) Ꞌ = 1 / (1 + x²). We have: yꞋ = ½ * (1 + x²) ½ – 1 * (1 + x²) Ꞌ * arctgx + √ (1 + x²) * (1 / (1 + x²)) = x * arctgx / √ (1 + x² ) + 1 / √ (1 + x²) = (x * arctgx + 1) / √ (1 + x²). Therefore, dy = ((x * arctgx + 1) / √ (1 + x²)) dx
Answer: dy = ((x * arctgx + 1) / √ (1 + x²)) dx.