How do you find the derivative of #G(x)=int (t^2-3t)dt# from #[1,x^2]#?

1 Answer
Jan 11, 2017

Answer:

# G'(x) = 2x^5-6x^3 #

Explanation:

We have:

# G(x) = int_1^(x^2) (t^2-3t) \ dt #

If we let

# F(x) = int_a^x (t^2-3t) \ dt #

Then from the first part of the Fundamental Theorem of Calculus, we have:

# F'(x) = x^2-3x " "....... [1] #

And from the second part of the Fundamental Theorem of Calculus, we have:

# int_a^x (t^2-3t) \ dt = F(x) - F(a) " "....... [2] #

Therefore,

# G(x) = F(x^2) - F(a) " "# from [2]

Differentiating wrt #x# we get

# \ \d/dx G(x) = d/dx F(x^2) - d/dx F(a) #
# :. G'(x) = F'(x^2)d/dx (x^2) - 0 " "# chain rule
# " "= ((x^2)^2-3(x^2)) (2x) " "# from [1]
# " "= (x^4-3x^2) (2x) #
# " "= 2x^5-6x^3 #