If y=x^ x +( x)^x , then d y d x at x= 2 is

Step-by-step solution

Given, y=x^ x +( x)^x Let u=x^ x and v=( x)^x Now, u=x^ x , u= x x Differentiating w.r.t. x , we get aligned & 1 u d u d x = x 1 x + x x & d u d x =u ( x x + x x ) & d u d x =x^ x ( x x + x x ) aligned Now, v=( x)^x v=x ( x) Differentiating w.r.t. x , we get 1 v d v d x =x x x + ( x) aligned & d v d x =v(x x+ x) & d v d x =( x)^x(x x+ x) aligned Adding Eqs. (i) and (ii), we get aligned & d u d x + d v d x =x^ x ( x x + .+ x x) &+( x)^x(x x+ x) & d y d x =x^ x ( x x + x x ) &++( x)^x(x x+ x) aligned At x= 2 aligned