# Differentiate the following functions with respect to x. Differentiating an integral?

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!

## Answers and Replies

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.

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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!