^{4}), t=0..x

^{1/3})

^{1/3}, then

^{4}), t=0..z)

^{-2/3}

^{4}) = sqrt (1+x

^{4/3})

^{4/3}) * (1/3)x

^{-2/3}

^{-2/3}, F'(0) is undefined.

Let F (x) = integral from 0 to x^(1/3) of function sq root (1 + t^4) dt. F'(0) = ?

Solution manual says: F (x) = f(g(x)), where g(x) = x^(1/3) and f (z) = integral from 0 to z of function sq root (1+t^4) dt. Then they apply chain rule. Do not understand

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Marked as Best Answer

We are given the function

F(x) = int( sqrt(1+t^{4}), t=0..x^{1/3})

We do not need to evaluate this integral, since we only want the derivative F'(x) and can therefore use the Fundamental Theorem of Calculus.

Use the substitution z=x^{1/3}, then

F(z(x)) = int( sqrt(1+t^{4}), t=0..z)

By the chain rule,

F'(x) = dF/dx = dF/dz * dz/dx

We have

dz/dx = (1/3)x^{-2/3}

and, by the Fundamental Theorem,

dF/dz = sqrt(1+z^{4}) = sqrt (1+x^{4/3})

so that

F'(x) = sqrt (1+x^{4/3}) * (1/3)x^{-2/3}

Because of the term x^{-2/3}, F'(0) is undefined.

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## Comments

of the derivative of a function is just that function (up to an additive constant) and the derivative of an integral of a function is that function. For indefinite integrals (antiderivatives) this statement is true by definition. For definite integrals, usually interpreted as “area under the graph” the statement is only true if we turn the integral into a function, by letting its upper limit be a variable, say z: then F(z) = int (f(t), t=0..z) is a new function, defined for all z where the integral exists.

Notice that the integration variable t here is only internal to the integral; the function F(z) does not depend on it. We could change the letter of the integration variable (e.g., from t to u) without changing the function F(z). This is why an integration variable is often referred to as a "dummy variable."

The Fundamental Theorem, in one of its many formulations, now says that F(z) is differentiable and

F’(z) = f(z)

To apply the theorem, wherever you see the integration variable t in the integrand, you need to change it to the free variable z. This eliminates t from the problem (answering your last question first).

Now what if the upper limit of the integral is not just z, but some more complicated function of some other variable x? The standard way of dealing with this is to define z to be this more complicated function, z=z(x).

This means that the integral is now a function of z, which itself is a function of x, F=F(z(x)). In cases where you have a composition of two functions, you need to use the chain rule to find the derivative with respect to x. Therefore,

F’(x) = dF/dz * dz/dx.

_{0}^{x}f(t) dt,