Step-by-step explanation:
Use the second fundamental theorem of calculus.
If y = ∫ₐᵇ f(t) dt, then dy/dx = f(b) db/dx − f(a) da/dx
dy/dx = e^(x³) tan (x³) (3x²) − e^(3x) tan (3x) (3)
dy/dx = 3x² e^(x³) tan (x³) − 3 e^(3x) tan (3x)
Answer:
3[x²(e^x³)(tan(x³) - (e^3x)tan(3x)]
Derivative of x³ = 3x²
Derivative of 3x = 3
3x²(e^x³)(tan(x³) - 3(e^3x)tan(3x)
y = kx
30 = 6k
k = 5
a) y = 5x
y = 5 × 12
b) y = 60
ok
, where a=x^3, b=3x
