If f(x)= g(x)+g(-x) 2 + 2 [h(x)+h(-x)]⁻¹ where g and h are differentiable function, then f ^ (0)

Step-by-step solution

Given, f(x)= g(x)+g(-x) 2 + 2 [h(x)+h(-x)]⁻¹ f(x)= g(x)+g(-x) 2 +2[h(x)+h(-x)] On differentiating w.r.t. x , we get f^ (x)= g^ (x)-g^ (-x) 2 +2 [h^ (x)-h^ (-x) ] f^ (0)= g^ (0)-g^ (0) 2 +2 [h^ (0)-h^ (0) ]=0