Leibniz's Rule (1 Viewer)

leehuan · May 29, 2016

I just want to check my method

$$Q: Use Leibniz's rule to differentiate both sides of the equation$\\ \int^x{\left(a^2+x^2\right)^{-\frac{1}{2}}dx}=\sinh^{-1}{\frac{x}{a}}+c\\$ with respect to $a$. Hence find a formula for $\int{\left(1+x^2\right)^{-\frac{3}{2}}dx}$

(The question put that superscript x on the integral... not sure if that was intended or a mistake... anyway)

$$By Leibniz's rule$\\ \begin{align*} \int{\frac{\partial}{\partial a} \left(a^2+x^2\right)^{-\frac{1}{2}}dx} &= \frac{d}{da} \left( \sinh^{-1}{\frac{x}{a}} + c\right)\\ \Leftrightarrow \int{-a \left(a^2+x^2 \right)^{-\frac{3}{2} } dx} &= -\frac{x}{a^2}\cdot \frac{1}{\sqrt{1+{\left(\frac{x}{a}\right)}^2}} \\ a=1: \\ \int{\left(1+x^2\right)^{-\frac{3}{2}} dx} &= \frac{x}{\sqrt{1+x^2}}\end{align*}$

Paradoxica · May 29, 2016

leehuan said:
I just want to check my method

$$Q: Use Leibniz's rule to differentiate both sides of the equation$\\ \int^x{\left(a^2+x^2\right)^{-\frac{1}{2}}dx}=\sinh^{-1}{\frac{x}{a}}+c\\$ with respect to $a$. Hence find a formula for $\int{\left(1+x^2\right)^{-\frac{3}{2}}dx}$

(The question put that superscript x on the integral... not sure if that was intended or a mistake... anyway)

$$By Leibniz's rule$\\ \begin{align*} \int{\frac{\partial}{\partial a} \left(a^2+x^2\right)^{-\frac{1}{2}}dx} &= \frac{d}{da} \left( \sinh^{-1}{\frac{x}{a}} + c\right)\\ \Leftrightarrow \int{-a \left(a^2+x^2 \right)^{-\frac{3}{2} } dx} &= -\frac{x}{a^2}\cdot \frac{1}{\sqrt{1+{\left(\frac{x}{a}\right)}^2}} \\ a=1: \\ \int{\left(1+x^2\right)^{-\frac{3}{2}} dx} &= \frac{x}{\sqrt{1+x^2}}\end{align*}$

Everything is fine.

leehuan · May 29, 2016

$$Let $x(t) = \int_0^t{e^{t-s}f(s)ds}.$ Use the chain rule, Leibniz's rule and the FTC to show that $\\ \frac{dx}{dt}-x(t)=f(t)$

I have no idea how to apply all three rules into this question...

(Later question that was easier to complete: $$Check your result by observing that $x(t)=e^t\int_0^t{e^{-s}f(s)ds}$ so that nobody uses it)

InteGrand · May 29, 2016

leehuan said:
$$Let $x(t) = \int_0^t{e^{t-s}f(s)ds}.$ Use the chain rule, Leibniz's rule and the FTC to show that $\\ \frac{dx}{dt}-x(t)=f(t)$

I have no idea how to apply all three rules into this question...

(Later question that was easier to complete: $$Check your result by observing that $x(t)=e^t\int_0^t{e^{-s}f(s)ds}$ so that nobody uses it)

$\noindent Here's a generalised form Leibniz rule with variable limits too that you'll want to use:$

https://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement .

$\noindent To obtain something like that (for your one the lower limit is a constant), essentially set the upper limit to be $u$. So we have $ x=F(u,t) = \int _0 ^u e^{t-s}f(s) \text{ d}s $. Then let $u \equiv u(t)=t$, then from the multivariable chain rule,$

$\begin{align*} \frac{ \mathrm{d}x}{\mathrm{d}t} &= \frac{ \partial F }{\partial u} \frac{\mathrm{d}u}{\mathrm{d}t} + \frac{ \partial F}{\partial t} \underbrace{ \frac{\mathrm{d}t}{\mathrm{d}t} }_{\text{just }1}. \end{align*}$

To compute these three terms, we'll need to use the various rules mentioned (FTC etc.). I'll let you have a go from here.

Paradoxica · May 29, 2016

leehuan said:
$$Let $x(t) = \int_0^t{e^{t-s}f(s)ds}.$ Use the chain rule, Leibniz's rule and the FTC to show that $\\ \frac{dx}{dt}-x(t)=f(t)$

I have no idea how to apply all three rules into this question...

(Later question that was easier to complete: $$Check your result by observing that $x(t)=e^t\int_0^t{e^{-s}f(s)ds}$ so that nobody uses it)

Hm, I'm getting dx/dt = f(t)

Clearly I'm doing something wrong, but I'm not quite sure what.

Paradoxica · May 29, 2016

InteGrand said:
$\noindent Here's a generalised form Leibniz rule with variable limits too that you'll want to use:$

https://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement .

$\noindent To obtain something like that (for your one the lower limit is a constant), essentially set the upper limit to be $u$. So we have $ x=F(u,t) = \int _0 ^u e^{t-s}f(s) \text{ d}s $. Then let $u \equiv u(t)=t$, then from the multivariable chain rule,$

$\begin{align*} \frac{ \mathrm{d}x}{\mathrm{d}t} &= \frac{ \partial F }{\partial u} \frac{\mathrm{d}u}{\mathrm{d}t} + \frac{ \partial F}{\partial t} \underbrace{ \frac{\mathrm{d}t}{\mathrm{d}t} }_{\text{just }1}. \end{align*}$

To compute these three terms, we'll need to use the various rules mentioned (FTC etc.). I'll let you have a go from here.

Ah, well that's fairly easy then, the result follows through almost immediately.

InteGrand · May 29, 2016

Paradoxica said:
Hm, I'm getting dx/dt = f(t)

Clearly I'm doing something wrong, but I'm not quite sure what.

Because the limits and integrand both depend on t, we can't just use FTC simply like that (I'm assuming you subbed t for s in the integrand to get your dx/dt).

leehuan · May 29, 2016

Paradoxica said:
Hm, I'm getting dx/dt = f(t)

Clearly I'm doing something wrong, but I'm not quite sure what.

I had the same problem lol. I forgot about the dependence-on-the-variable problem

leehuan · May 29, 2016

I'm dumb. Does du/dt = 1 when we let u=t?

InteGrand · May 29, 2016

leehuan said:
I'm dumb. Does du/dt = 1 when we let u=t?

Yeah.

leehuan · Jun 11, 2016

Forgotten how to do this

$\\$Suppose $a>-1, b>-1.$ Use Leibniz's rule to show that $\\\int_0^1{\frac{x^a-1}{\ln{x}}dx}=\ln{(a+1)}$

turntaker · Jun 11, 2016

Paradoxica said:
Everything is fine.

I agree

InteGrand · Jun 11, 2016

leehuan said:
Forgotten how to do this

$\\$Suppose $a>-1, b>-1.$ Use Leibniz's rule to show that $\\\int_0^1{\frac{x^a-1}{\ln{x}}dx}=\ln{(a+1)}$

$\noindent Let $h(a)=\int _{0} ^{1} \frac{x^a-1}{\ln{x}} \text{ d}x $. Applying Leibniz' Rule, $h^{\prime}\left(a\right) =\int _{0} ^{1} \frac{\partial}{\partial a} \left(\frac{x^a-1}{\ln{x}}\right) \text{ d}x = \int _0 ^1 x^{a}\text{ d}x = \frac{1}{a+1}$, since $\frac{\partial}{\partial a} \left(\frac{x^a-1}{\ln{x}}\right)=\frac{\ln x \cdot x^{a}}{\ln x}=x^{a}$. So $h(a) = \ln \left(a+1\right)+C$, for some constant $C$. But $h(0)=0$, as $h(0) = \int _0 ^1 \frac{1-1}{\ln x}\text{ d}x=0$. So $\ln \left(0+1\right)+C = 0 \Rightarrow C = 0$. Thus $h(a)=\ln \left(a+1\right)$, as desired.$

leehuan · Jun 11, 2016

Gonna revive this one temporarily

InteGrand said:
$\noindent Here's a generalised form Leibniz rule with variable limits too that you'll want to use:$

https://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement .

$\noindent To obtain something like that (for your one the lower limit is a constant), essentially set the upper limit to be $u$. So we have $ x=F(u,t) = \int _0 ^u e^{t-s}f(s) \text{ d}s $. Then let $u \equiv u(t)=t$, then from the multivariable chain rule,$

$\begin{align*} \frac{ \mathrm{d}x}{\mathrm{d}t} &= \frac{ \partial F }{\partial u} \frac{\mathrm{d}u}{\mathrm{d}t} + \frac{ \partial F}{\partial t} \underbrace{ \frac{\mathrm{d}t}{\mathrm{d}t} }_{\text{just }1}. \end{align*}$

To compute these three terms, we'll need to use the various rules mentioned (FTC etc.). I'll let you have a go from here.

Back then, using the multivariable chain rule I did get the answer out successfully, but once again I'm stumped on technicality.

We let x=F(u,t), which that step I do get

But is that really x(t)? Or is it x(something else)?

Or does it forcibly become x(t) the instant we let u=t

Paradoxica · Jun 11, 2016

leehuan said:
Forgotten how to do this

$\\$Suppose $a>-1, b>-1.$ Use Leibniz's rule to show that $\\\int_0^1{\frac{x^a-1}{\ln{x}}dx}=\ln{(a+1)}$

Differentiate with respect to a

leehuan · Jun 11, 2016

Paradoxica said:
Differentiate with respect to a

I messed that one up cause for some reason whilst I wrote partial/partial a I still tried to treat it as a polynomial and not an exponential lol

leehuan · Jun 15, 2016

Once again please check my working

$$The function $h(x)$ is given by $h(x)=\int_0^{\cosh{x}}{k(x+t)dt}\\ $where $k$ is continuously differentiable.$ \\ $Express $h^\prime$ in terms of $k$

Answer (some steps/reasons omitted just for ease of typing):
$$Let $y=H(x,u)=\int_0^{\cosh{u}}{k(x+t)dt}\\ \frac{\partial }{\partial u}\int_0^{\cosh{u}}{k(x+t)dt} =\sinh{u} \cdot k(x+\cosh{u})$

$\frac{\partial }{\partial x}\int_0^{\cosh{u}}{k(x+t)dt} = \int_0^{\cosh{u}}{ \frac{\partial }{\partial x} k(x+t) dt} = k(x+t)|_0^{\cosh{x}}$

$\frac{dy}{dx}= \frac{\partial f}{\partial x} \frac{dx}{dx}+ \frac{\partial f}{\partial u}\frac{du}{dx}$

$$When $u=x\\ \frac{dy}{dx}=\frac{d}{dx}h(x)=h^\prime (x) \\ = \sinh{x} \cdot k(x+\cosh{x}) + k(\cosh{x}+x) - k(x)$

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