If y = (x^2 + 1)^ x for x > 0 such that dy dx = y [ 2x x g(x) + x [g(x)] ] , then the function 1 g(x) is...

Step-by-step solution

Given y = (x^2 + 1)^ x Taking natural logarithm on both sides: y = x (x^2 + 1) Differentiating with respect to x : 1 y dy dx = x (x^2 + 1) + x 2x x^2 + 1 dy dx = y [ 2x x x^2 + 1 + x (x^2 + 1) ] Comparing with the given expression, we get g(x) = x^2 + 1 . Let h(x) = 1 g(x) = 1 x^2 + 1 . Differentiating h(x) with respect to x : h'(x) = -2x (x^2 + 1)^2 For x > 0 , h'(x) Thus, the function 1 g(x) is strictly decreasing for x > 0 .