What is the Derivative of sin x^5?

  • Thread starter Thread starter char808
  • Start date Start date
  • Tags Tags
    Derivative Sin
Click For Summary
SUMMARY

The derivative of the integral of sin(t) from 0 to x^5 is calculated using the Leibniz rule for differentiation. The correct result is 5x^4 sin(x^5), which involves first integrating sin(t) with respect to t and then applying the limits of integration. The confusion arose from the notation used in the original post, where 'o' was mistakenly used instead of '0'. Understanding the application of the Leibniz rule is crucial for differentiating integrals with variable limits.

PREREQUISITES
  • Understanding of calculus concepts, specifically differentiation and integration.
  • Familiarity with the Leibniz rule for differentiating integrals.
  • Knowledge of LaTeX for mathematical notation.
  • Basic understanding of trigonometric functions, particularly sin(t).
NEXT STEPS
  • Study the Leibniz rule for differentiating integrals in depth.
  • Practice integrating and differentiating various functions using LaTeX.
  • Explore advanced applications of the Leibniz rule in complex integrals.
  • Review common mistakes in mathematical notation, particularly in calculus.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on calculus, as well as anyone looking to improve their understanding of differentiation techniques involving integrals.

char808
Messages
27
Reaction score
0
Find the derivative.

d/dx \int\stackrel{x^5}{o} sint dt

I came up with sin x^5.

The answer is 5x^4 sin (x^5).



Not sure what I'm missing here.
 
Last edited:
Physics news on Phys.org
Is the question:

Integrate x^5*sin(t) with respect to 't', then differentiate the result with respect to 'x'?
 
x^5 and 0 are the limits, right?

First: you integrate sint with respect to t.
Second: Apply the limits to the value you obtained for the integer. You'll get two terms, get it?

Then, derive this result with respect to x.
 
emol1414 said:
x^5 and 0 are the limits, right?

First: you integrate sint with respect to t.
Second: Apply the limits to the value you obtained for the integer. You'll get two terms, get it?

Then, derive this result with respect to x.

That works in this problem but the point is likely to use Leibnitz rule for differentiating an integral with variable upper limit. It would work even if you had something more difficult than sin(t) that you couldn't find the antiderivative for. The basic Leibnitz rule is:

\frac d {dx}\int_a^{g(x)} f(t)\, dt = f(g(x))\cdot g'(x)
 
The repaired LaTeX is below.
char808 said:
Find the derivative.

d/dx \int\stackrel{x^5}{o} sint dt

I came up with sin x^5.

The answer is 5x^4 sin (x^5).



Not sure what I'm missing here.

d/dx \int_0^{x^5} sint dt
To the OP: click the integral to see the LaTeX code I used.
The integral in the original post was confusing to at least one person who didn't understand that x5 was one of the limits of integration.

Also, don't use o (the letter) where 0 (the numeral) is intended.

BTW, there is no such word as "intergral" - the word is integral.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K