If (x) = x^3 ₂^x ( 5 t^2 - 2 '(t) ) dt , x > 0 , and the tangent to the curve y = (x) at x = 2 passes through the point ( , 10) , then the value of is equal to

Step-by-step solution

Given equation is: (x) = x^3 ₂^x ( 5 t^2 - 2 '(t) ) dt First, find the value of the function at x = 2 : (2) = 2^3 ₂^2 ( 5 t^2 - 2 '(t) ) dt = 0 So, the point of tangency is (2, 0) . Now, differentiate the given equation with respect to x using the product rule and Leibniz rule: '(x) = 3x^2 ₂^x ( 5 t^2 - 2 '(t) ) dt + x^3 ( 5 x^2 - 2 '(x) ) Substitute x = 2 to find the slope of the tangent: '(2) = 0 + 2^3 ( 5 2^2 - 2 '(2) ) '(2) = 8 ( 5 4 - 2 '(2) ) '(2) = 10 - 16 '(2) 17 '(2) = 10 '(2) = 10 17 The equation of the t