Differentiating Integrals w/ Respect to x: Tips & Tricks

[LilTaru]

Homework Statement

Differentiate the following functions with respect to x.

a) \int (t² + cos(t⁷))/(1 + t⁴) dt [0, x]

b) \int (t² + cos(t⁷))/(1 + t⁴) dt [2, x]

c) \int (t² + cos(t⁷))/(1 + t⁴) dt [0, sin x]

d) \int (t² + cos(t⁷))/(1 + t⁴) dt [x, sin x]

e) \int (x² + cos(t⁷))/(1 + x⁴) dt [2, sin x]

Note: [a, b] means a is the lower limit of the integral and b is the upper limit of the integral.

Homework Equations

I know that the derivative of an indefinite integral is the function itself and that with definite integrals you need to find an antiderivative G(x) and the derivative equals G(upper limit) - G(lower limit).

The Attempt at a Solution

I have no idea how to find an antiderivative for this function. Nor do I know how to use sin(x) as an upper limit. Nor do I know how to do (e) with x replacing some of the t's. Will x be considered a constant then, since we are differentiating in respect to t? I don't want the answers to all of them... I prefer to do the work on my own... That's how I actually learn, but if someone can give me a hint on how to start each one? I would be greatly appreciative!

 

[armolinasf]

They're asking you to apply the second fundamental theorem of calculus which allows you to sort of skip finding an antiderivative. The logic goes something like this:

according to the first fundamental theorem, each of those integrals can be rewritten as the difference of their antiderivaties evaluated at the bounds, that is F(b)-F(a)

But you're asked to find the derivative of that, so you have d/dx(F(b)-F(a)) which essentially undoes the anti-differentiating and you're left with F'(b)-F'(a) which is just the function written inside the original integral.

In the case where you use sinx as an upper limit and say some constant c as a lower you have d/dx(F(sinx)-F(c)) Remember, F(sinx) is a composition of two functions and to differentiate it you need to apply the chain rule, also F(c) is a constant so its derivative is zero. In the end you're left with: F'(sinx)*cosx-0.

So the general form for problems like these is [F'(g(x))*g'(x)]-[F'(h(x))-h'(x)], where g and h are your upper and lower bounds respectively and F' is the function inside the original integral.

Hope this helps.

 

[LilTaru]

Oh. So I just differentiate the actual function? That sounds a lot easier. To clarify: when sin x is a limit, I plug in sin x into f and then differentiate the function, then multiply all of that by cos x? And in (e) do we treat the x's in the function as a constant?

 

[LilTaru]

Or do I differentiate f first and plug sin x into the already differentiated function?

 

[armolinasf]

f would already be differentiated so in (d) you would have (sinx^2 + cos(sinx^7))/(1 + sinx4) all of that times cosx then minus (x2 + cos(x7))/(1 + x4) all of that times one.

As far as treating x as a constant, I would probably agree with you. Otherwise you have a multivariable derivative. I would imagine your book would specify

 

[LilTaru]

Oh, okay! This makes a lot of sense now! Thank you so much!