MHB Can the Derivative of the Function $xe^x\csc{x}$ Be Solved Using the Chain Rule?

[Petrus]

Hello,
I got A problem i got hard with solving i am going to derivate/defferentiate $xe^x\csc{x}$ and what i think how you derivate it and get $f('x)=(x+1)e^x-xe^x\csc{x} \cot x$

edit: i use chain rule twice first on \(xe^x\) and then on the function. Hope you guys understand how i am thinking.

 

[chisigma]

Hello,
I got A problem i got hard with solving i am going to derivate/defferentiate $xe^xcscx$ and what i think how you derivate it and get $f('x)=(x+1)e^x-xe^xcscxcotx$

edit: i use chain rule twice first on xe^x and then on the function. Hope you guys understand how i am thinking.

What I undestand is that You have to solve the differential equation...

$\displaystyle f^{\ '}(x) = e^{x}\ (1+ x - x\ \csc x\ \cot x)$ (1)

... isn't it?...

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

Hello,
I got A problem i got hard with solving i am going to derivate/defferentiate $xe^xcscx$ and what i think how you derivate it and get $f('x)=(x+1)e^x-xe^xcscxcotx$

edit: i use chain rule twice first on xe^x and then on the function. Hope you guys understand how i am thinking.

Hi Petrus. :)

You are using the product rule and not the chain rule.
$$(u\cdot v)' = u' \cdot v + u \cdot v'$$
What you have is almost correct!
There is something missing though.

You'd get:
$$(x e^x \csc x)' = (x e^x)' \cdot \boxed{\csc x} + x e^x \cdot (\csc x)'$$
Where did you leave the $\boxed{\csc x}$ in the first term?

 

[soroban]

Hello, Petrus!

Differentiate $y \:=\:xe^x\csc x$

I got: $y'\:=\: (x+1)e^x-xe^x\csc x\cot x$ . . . . not quite

$\text{We have the product of }three\text{ functions: }\:y \:=\:\underbrace{x}_f\cdot \underbrace{e^x}_g\cdot \underbrace{\csc x}_h$

$\text{There is an Extended Product Rule . . .}$

. . $\text{If }\,y \:=\:f(x)\!\cdot\!g(x)\!\cdot\!h(x)$

. . $\text{then: }\,y' \:=\:f'(x)\!\cdot\!g(x)\!\cdot\!h(x) + f(x)\!\cdot\!g'(x)\!\cdot\!h(x) + f(x)\!\cdot\!g(x)\!\cdot\!h'(x)$$\text{Therefore: }\:y' \;=\;1\!\cdot\!e^x\!\cdot\!\csc x + x\!\cdot\!e^x\!\cdot\!\csc x + x\!\cdot\!e^x\!\cdot\!(\text{-}\csc x\cot x)$

. . . . . . . .$y' \;=\;e^x\csc x + xe^x\csc x - xe^x\csc x\cot x$

. . . . . . . .$y' \;=\;e^x\csc x(1 + x - x\cot x)$

 

[Petrus]

Thanks for all help. Sorry i did forget to write cscx
$f('x)=(x+1)ecscx-xe^xcscxcotx$

 

[Ackbach]

Use the backslash for a command for a trig function, followed by parentheses and the argument, like this:

$ \cos( \theta)$

yields $ \cos( \theta)$. It looks much better.

 

[Petrus]

Hello, Petrus!


$\text{We have the product of }three\text{ functions: }\:y \:=\:\underbrace{x}_f\cdot \underbrace{e^x}_g\cdot \underbrace{\csc x}_h$

$\text{There is an Extended Product Rule . . .}$

. . $\text{If }\,y \:=\:f(x)\!\cdot\!g(x)\!\cdot\!h(x)$

. . $\text{then: }\,y' \:=\:f'(x)\!\cdot\!g(x)\!\cdot\!h(x) + f(x)\!\cdot\!g'(x)\!\cdot\!h(x) + f(x)\!\cdot\!g(x)\!\cdot\!h'(x)$$\text{Therefore: }\:y' \;=\;1\!\cdot\!e^x\!\cdot\!\csc x + x\!\cdot\!e^x\!\cdot\!\csc x + x\!\cdot\!e^x\!\cdot\!(\text{-}\csc x\cot x)$

. . . . . . . .$y' \;=\;e^x\csc x + xe^x\csc x - xe^x\csc x\cot x$

. . . . . . . .$y' \;=\;e^x\csc x(1 + x - x\cot x)$

Hello Soroban,
I wanted to say i am really proud that you told me that! That is something really good to know that I never thought about. May I ask is that something should figure out by myself? I honestly just thought about with two functions! I say it again THANKS for sharing that idé