MHB Chain Rule & Trig Functions: Explaining to DeusAbs

[DeusAbscondus]

this time it relates to the chain rule and trig functions.

for example:
$$y=ln(sinx)\Rightarrow y'=cot(x)$$ via the chain rule, including the following steps
$$\frac{d}{dx}ln(sinx)$$
$$=cos(x).\frac{1}{sin(x)}$$
$$=\frac{cos(x)}{sin(x)}$$
$=cot(x)$
but I would have expected that, following the chain rule, an operation was required on the $sin(x)$, but it just gets shunted down into the denominator along with $x$ Could someone kindly explain to me why?

I mean I understand this expression: $$y=ln(x)\Rightarrow y'=\frac {1}{x}$$
but how does the sine function make it down into the denominator holus-bolus in my example above?

I'm sorry for the awkwardness of expression.
Can anyone see the difficulty I am having with this?

Thanks,
DeusAbs

 

[CaptainBlack]

this time it relates to the chain rule and trig functions.

for example:
$$y=ln(sinx)\Rightarrow y'=cot(x)$$ via the chain rule, including the following steps
$$\frac{d}{dx}ln(sinx)$$
$$=cos(x).\frac{1}{sin(x)}$$
$$=\frac{cos(x)}{sin(x)}$$
$=cot(x)$
but I would have expected that, following the chain rule, an operation was required on the $sin(x)$, but it just gets shunted down into the denominator along with $x$ Could someone kindly explain to me why?

I mean I understand this expression: $$y=ln(x)\Rightarrow y'=\frac {1}{x}$$
but how does the sine function make it down into the denominator holus-bolus in my example above?

I'm sorry for the awkwardness of expression.
Can anyone see the difficulty I am having with this?

Thanks,
DeusAbs

Put \(u=\sin(x)\), so \(y=\ln(u)\), then:

\[\frac{dy}{dx}=\frac{dy}{du}\;\frac{du}{dx}=\left( \frac{1}{u} \right) \left( \cos(x) \right) =\frac{\cos(x)}{\sin(x)}\]

CB

 

[DeusAbscondus]

and triple application of rule

Thanks Cap'n.

I think I'm on a roll: following on from what you and SuperSonic and others have shared about setting out my work, I'm going to try applying the chain rule triply, to the following:
$f(x)=ln(cos(x^2))\Rightarrow f'(x)=-2xtan(x^2)$
with the working out thus: (please comment if I can omit, add anything to or change anything about the way I set this out):
$Let$
$u=x^2$
$v=cos(u)$
$y=ln(v)$
and
$u'=2x$
$v'=-sin(x^2)$
$y'=\frac{1}{cos(x^2)}$
$\therefore f'(x)=u'.v'.y'=\frac{-2x.sin(x^2}{cos(x^2)}=-2xtan(x^2)$

Thanks for the great help,
DeusAbs

 

[SuperSonic4]

Re: and triple application of rule

Thanks Cap'n.

I think I'm on a roll: following on from what you and SuperSonic and others have shared about setting out my work, I'm going to try applying the chain rule triply, to the following:
$f(x)=ln(cos(x^2))\Rightarrow f'(x)=-2xtan(x^2)$
with the working out thus: (please comment if I can omit, add anything to or change anything about the way I set this out):
$Let$
$u=x^2$
$v=cos(u)$
$y=ln(v)$
and
$u'=2x$
$v'=-sin(x^2)$
$y'=\frac{1}{cos(x^2)}$
$\therefore f'(x)=u'.v'.y'=\frac{-2x.sin(x^2}{cos(x^2)}=-2xtan(x^2)$

Thanks for the great help,
DeusAbs

Nothing wrong there :)

 

[DeusAbscondus]

Re: and triple application of rule

Nothing wrong there :)

Ah, well that'd be because I try to model myself on you, Super.
D'abs

 

[CaptainBlack]

Re: and triple application of rule

Thanks Cap'n.

I think I'm on a roll: following on from what you and SuperSonic and others have shared about setting out my work, I'm going to try applying the chain rule triply, to the following:
$f(x)=ln(cos(x^2))\Rightarrow f'(x)=-2xtan(x^2)$
with the working out thus: (please comment if I can omit, add anything to or change anything about the way I set this out):
$Let$
$u=x^2$
$v=cos(u)$
$y=ln(v)$
and
$u'=2x$
$v'=-sin(x^2)$
$y'=\frac{1}{cos(x^2)}$
$\therefore f'(x)=u'.v'.y'=\frac{-2x.sin(x^2}{cos(x^2)}=-2xtan(x^2)$

Thanks for the great help,
DeusAbs

I think you would be better off with Liebniz' notation rather that Newton's Fluxion like notation.

Then:

\[\frac{d}{dx}f(x)=\frac{df}{dv}\frac{dv}{du}\frac{du}{dx}\]

which is much more suggestive of what is going on.

CB

 

[DeusAbscondus]

Re: and triple application of rule

I think you would be better off with Liebniz' notation rather that Newton's Fluxion notation.

Then:

\[\frac{d}{dx}f(x)=\frac{df}{dv}\frac{dv}{du}\frac{du}{dx}\]

which is much more suggestive of what is going on.

CB

Thanks Cap'n; I'll look it up and look into it. "Fluxion" sure sounds alchemical. If I could find a way of doing Calculus sans any mystical overtones, so much the better for a proudly upright atheist and scientific materialist.
D'abs