# Taking the derivative of some function

I define $$y =\exp\left(\int^x f(t)w(t) dt \right)$$ and i want to take the derivative with respect to x. Can I just say y'= $$f(x)w(x) \exp \left(\int^x f(t) w(t) dt\right)$$?

Consequently can I say that y'' = $$\left( f'(x)w(x)+f(x)w'(x) \right) \exp\left(\int^x f(t) w(t) dt\right) +f(x)w(x)f(x)w(x)\exp\left(\int^x f(t) w(t) dt\right)=$$ $$\left( f'(x)w(x)+f(x)w'(x) \right) \exp\left(\int^x f(t) w(t) dt\right) +f(x)^2 w(x)^2 \exp\left(\int^x f(t) w(t) dt\right)$$

Generally, what is the difference between $$\int^x f(t)w(t)dt$$ and just $$\int f(x)w(x)dx$$

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What are aaaaaaalllll the lower limits of the above integrals?! – DonAntonio Dec 2 '12 at 17:23
Is it just notation ? I dont understand it completely. I assume its just the indefinite integral with respect to x? – Applied mathematician Dec 2 '12 at 17:26
perhaps it is "just notation", but it might be important, depending on the functions involved. Anyway, I think the lower limtis are lacking. – DonAntonio Dec 2 '12 at 17:30
If your author means $$\int^xf(t)w(t)dt=\int_0^{x}f(t)w(t)dt$$ everything works out. (except for the second derivative, I think) – 000 Dec 2 '12 at 17:47
My mistake. Your second derivative is also correct! :) – 000 Dec 2 '12 at 17:56

The definition $$x\mapsto y(x):=\exp\left(\int^x f(t)w(t) dt \right)$$ defines the function $y(\cdot)$ only up to a multiplicative constant $\ne0$. In fact, the right side has to be interpreted as follows: Assume that $t\mapsto F(t)$ is an arbitrary primitive of $t\mapsto f(t)w(t)$. Then $y(x):=\exp\bigl(F(x)\bigr)$. Two such primitives $F_1$ and $F_2$ differ by an additive constant $c\in{\mathbb R}$, therefore two candidates $x\mapsto y(x)$ satisfying your definition differ by a multiplicative constant $C:=e^c\ne 0$.

From $y(x)=\exp\bigl( F(x)\bigr)$ we get $$y'(x)=\exp\bigl( F(x)\bigr)\cdot F'(x)=y(x)\cdot f(x)w(x)$$ and then $$y''(x)=y(x)\bigl(f^2(x)w^2(x)+f'(x)w(x)+f(x)w'(x)\bigr)\ .$$ The "arbitrary multiplicative constant" is still present on the right side of the last equation, but the result is written in such a way that after this constant has been chosen once and for all no further ambiguity persists.

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Are you referring to $\int_{0}^{x}f(t)w(t)dt$ and $\int f(x)w(x)dx$? Well, if we differentiate the first and second, we arrive at $f(x)w(x)$, but for conceptually different reasons.

We know that $$\frac{d}{dx}\int_{0}^{x}f(t)w(t)dt=f(x)w(x)$$

due to the following proof (part of the Fundamental Theorem of Calculus):

Let $u(t)=f(t)w(t)$. Define $U(t)=\int u(t)dt$. From this, we have \begin{align} \frac{d}{dx}\int_0^{x}f(t)w(t)dt&=\frac{d}{dx}\int_0^{x}u(t)dt\\ &=\frac{d}{dx}\left[ \int u(t)dt\right]^{x}_{0}\\ &=\frac{d}{dx}\left[U(x)-U(0)\right]\\ &=\frac{d}{dx}U(x)+\frac{d}{dx}U(0)\\ &=\frac{d}{dx}U(x)+0\\ &=u(x)\\ &=f(x)w(x). \end{align}

We know that $$\frac{d}{dx}\int f(x)w(x)dx=f(x)w(x)$$ on the basis of the fact that differentiation and integration are inverse operations just like addition and subtraction. They cancel one another, in other words.

Let's assume the author means $$y=\exp\left(\int_0^{x}f(t)w(t)dt\right).$$

Make the $u$ sub as above and say more aptly $$y=\exp\left(\int_0^{x}u(t)dt\right).$$

When we differentiate this, we must apply the chain rule: $$\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{dx}.$$

Let $v=\int_0^{x}u(t)dt$. We then have:

\begin{align} \frac{dy}{dx}&=\exp(v)\frac{d}{dx}v\\ &=\exp\left({\int_0^x u(t)dt}\right)\frac{d}{dx}\int_0^{x}u(t)dt\\ &=\exp\left({\int_0^x u(t)dt}\right)u(x)\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f(x)w(x). \end{align}

## Second Derivative

\begin{align} y''&=\left[\exp\left({\int_0^x u(t)dt}\right)u(x) \right]'\\ &=\left(\exp\left({\int_0^x u(t)dt}\right)\right)'u(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x) \quad (\text{product rule: } (fg)'=f'g+fg'.)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u(x)u(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u^2(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u^2(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f^2(x)w^2(t)+\exp\left({\int_0^x f(t)w(t)dt}\right)(f(t)w(t))'\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f^2(x)w^2(t)+\exp\left({\int_0^x f(t)w(t)dt}\right)(f(t)'w(t)+f(t)w'(t)). \end{align}

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When one writes $$\int g(x)dx,$$ the usual meaning is "an antiderivative of g".

When one writes $$\int_a^bg(x)dx,$$ one means the "integral of g on $[a,b]$" (or "definite integral"), which is a number defined using a limit of Riemann sums.

If one writes $$\int_a^xg(t)dt,$$ then for each $x$ this gives some number (i.e., for each $x$ one gets a definite integral, that is a number) and so we have a function on $F(x)$. The Fundamental Theorem of Calculus states the fact that in this last case $F(x)$ is an antiderivative for $f(x)$.

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and what does $$\int^x f(t) dt$$ mean? is it the same as F(X) ? – Applied mathematician Dec 2 '12 at 17:37
Probably. But it's a very unusual notation. – Martin Argerami Dec 2 '12 at 17:46

If the functions $\,f,w\,$ are such that $\,f(t)w(t)\,$ has a primitive $\,F(t)\,$ in some interval containing $\,x\,$ , say in $\,[a,x]\,$ , then the FTC tells us that

$$\int_a^xf(t)w(t)\,dt=F(x)-F(a)$$ so

$$y=e^{\int_a^xf(t)w(t)\,dt=F(x)-F(a)}\Longrightarrow \frac{dy}{dx}=y'=F'(x)\,e^{F(x)-F(a)}=f(x)w(x)\,e^{F(x)-F(a)}$$

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