Step by step
The question
Question: find the derivative of ∫ from 1 to x³ of 1/(t² + 1) dt.
ddx∫ from 1 to x³ of 1t2 + 1 dt =
1 The rule
The rule: the derivative of the area from a to x is the function at x.
Just put x in place of t.
ddx∫ from a to x of f(t) dt = f(x)
2 Put the top in place of t
The top is not just x: it is x³.
Call it u.
Its slope is u′.
top: u = x3, u′ = 3x²
The chain rule: put u in place of t, then multiply by u′.
ddx∫ from 1 to x³ of 1t2 + 1 dt = f(u) × u′
Put x³ in place of t, times 3x².
= 1x6 + 1 × 3x²
Careful!
Do not stop at f of u.
The top is x³, so we must multiply by its slope, 3x².
Careful: not 1x6 + 1 ✗
Write the 3x² in front: 3x² over (x⁶ + 1).
= 3x²x6 + 1
3 Check
Let's check at x = 2.
Our answer gives 0.1846.
answer at x = 2: 0.1846
Now the slope of the area function F between x = 2 and 2.001, with a calculator: about 0.184.
The same!
Right!
F(2.001) − F(2)0.001 ≈ 0.184
Let's look back.
First: the rule.
Then: put the top in place of t.
Last: check.
Answer
= 3x²x6 + 1