Find the derivative of the integral from 0 to x^4 (sec(t)dt)

find the derivative of the integral from 0 to x^4 (sec(t)dt)

I get sec(x^4)(4x^3)

I understand the procedure for solving this integral. However, I don't understand why exactly we need to use the chain rule in this situation.

Why isn't the answer just sec(x^4)?

Thank you!
 
[tex] \frac{d}{dt} \int_{0}^{x^{4}} \sec t dt [/tex]. Basically you are getting a derivative. If its a derivative then you have to use the chain rule.
 
Last edited:
Thanks for your reply. How about this problem

d/dx integral 0 to ln(x) (sint + t^3/2)dt

using the chain rule and ftc1 I get:

u = ln(x) du/dx = 1/x

d/du (sinu + u^3/2)du/dx

sin(lnx)+(lnx)^3/2 (1/x)

is this correct? Thanks again.
 
[tex] \frac{d}{dx} \int_{0}^{\ln x} (\sin t + t^{\frac{3}{2}}) dt [/tex]. So the derivative so far, is [tex] \sin(\ln x) + (\ln x)^\frac{3}{2} dt [/tex]. So then it would be:

[tex] \frac{\sin(\ln x)}{x} + \frac{(\ln x)^{\frac{3}{2}}}{x} [/tex].
 
Last edited:
so my answer was off by not multiplying both parts by 1/x? Thanks.
 
Last edited:
I took the derivative of u. i confirmed your answer.
 
ok ok, thanks!!
 
yeah, both parts need to be multiplied by [tex] \frac{1}{x} [/tex]. They both contain u and thus must contain du .
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top