The derivative of e^ a x b x with respect x is r^ a x b x ⁻¹ b a . When a>0, b>0 , then value of r , is

Step-by-step solution

Given, d dx ( e ^ ax bx )= re ^ ax ( bx + ) , then r = ? Let y=e^ a x b x d y d x =a e^ a x b x-b e^ a x b x d y d x =e^ a x (a b x-b b x) Let . array l a = r b = r array ...(i) Then, d y d x =e^ a x r b x - b x d y d x =e^ a x r (b x+ ) ...(ii) Where, = b a = ⁻¹ ( b a ) and aligned r ² ( ² + ² ) &= a ²+ b ² r ² &= a ²+ b ² r &= a ²+ b ² aligned