Homework question involving 2nd FTC, derivative of an integral

[mrchris]

Homework Statement

Suppose that the function f : R → R is differentiable. Define the function H: R → R by
H(x) = ∫−x to x of [ f (t) + f (−t)]dt for all x in R.
Find H'' (x).

Homework Equations

The Attempt at a Solution

i divided it up into ∫ 0 to x of [ f (t) + f (−t)]dt and -∫ from 0 to -x of [ f (t) + f (−t)]dt. then by the 2nd FTC, H '(x)=
2[ f (x) + f (−x)], so H ''(x) should then be of 2[ f '(x) - f '(−x)]. I'm just not sure if the f(-t) changes anything besides the fact that when I differentiate f(-t) I use the chain rule and multiply by -1.

 

[Mark44]

Homework Statement

Suppose that the function f : R → R is differentiable. Define the function H: R → R by
H(x) = ∫−x to x of [ f (t) + f (−t)]dt for all x in R.
Find H'' (x).

Homework Equations

The Attempt at a Solution

i divided it up into ∫ 0 to x of [ f (t) + f (−t)]dt and -∫ from 0 to -x of [ f (t) + f (−t)]dt.

Looks OK up to here.

So $$ H(x) = -\int_0^{-x}(f(t) + f(-t))dt + \int_0^{x}(f(t) + f(-t))dt $$

Show me how you got H'(x).

then by the 2nd FTC, H '(x)=
2[ f (x) + f (−x)], so H ''(x) should then be of 2[ f '(x) - f '(−x)]. I'm just not sure if the f(-t) changes anything besides the fact that when I differentiate f(-t) I use the chain rule and multiply by -1.

 

[mrchris]

so i am under the impression that d/dx of ∫f(t) dt = f(x), so taking each term, H'(x)= dH/dx of
∫0 to x [f(t)]= f(x)
∫0 to x [f(-t)]= f(-x)
∫0 to -x [f(t)]= f(-x)*-1
∫0 to -x [f(-t)]= f(x)*-1

so H'(x)= [f(x)+f(-x)]-(-1)[f(-x)+f(x)]=[f(x)+f(-x)]+[f(-x)+f(x)]=2[f(x)+f(-x)]

I am not sure if I am also supposed to be applying the chain rule at this step with the f(-t) derivative

 

[Mark44]

This looks fine. Now take the derivative again, keeping in mind that d/dx(f(-x)) = -f'(-x).

 

[mrchris]

so H ''(x) should then be of 2[ f '(x) - f '(−x)].

 

[HallsofIvy]

so i am under the impression that d/dx of ∫f(t) dt = f(x), so

You mean, I hope, that [itex]d/dx \int_a^x f(t)dt[/itex] so that [itex]d/dx \int_x^a f(t)dt= d/dx\left(-\int_a^x f(t)dt\right)= -f(x)[/itex]

taking each term, H'(x)= dH/dx of
∫0 to x [f(t)]= f(x)
∫0 to x [f(-t)]= f(-x)
∫0 to -x [f(t)]= f(-x)*-1
∫0 to -x [f(-t)]= f(x)*-1

so H'(x)= [f(x)+f(-x)]-(-1)[f(-x)+f(x)]=[f(x)+f(-x)]+[f(-x)+f(x)]=2[f(x)+f(-x)]

I am not sure if I am also supposed to be applying the chain rule at this step with the f(-t) derivative