MHB How can I find the derivative of ln(sin5x)?

[DeusAbscondus]

Hi folks,
For some reason, the following does not make sense to me:

If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$

Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs

 

[Sudharaka]

Hi folks,
For some reason, the following does not make sense to me:

If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$

Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs

Hi DeusAbscondus, :)

What you need here is the Chain rule of differentiation. Please refer the link given and try to understand how to use this rule so that you can easily differentiate the given function.

\[f(x)=\ln(\sin(5x))\]

Using the chain rule of differentiation we can write,

\[\frac{d}{dx}f(x)=\frac{d}{d(\sin(5x))}f(x)\frac{d}{dx}\sin(5x)\]

Try to continue from here. :)

Kind Regards,
Sudharaka.

 

[CaptainBlack]

Hi folks,
For some reason, the following does not make sense to me:

If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$

Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs

The derivative of \(\sin(5x)\) is \(5 \cos(5x)\) not \( 5\cos(x)\).CB

 

[SuperSonic4]

Hi folks,
For some reason, the following does not make sense to me:

If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$

Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs

I'll do this by substitution to make it easier to see. On an exam you'd be fine to just go with it.

Let

• $u(x) = 5x$
• $v(u) = \sin(u)$
• $y(v) =\ln(v)$

Their respective derivatives are

Spoiler

• $\dfrac{du}{dx} = 5$
• $\dfrac{dv}{du} = \cos(u)$
• $\dfrac{dy}{dv} = \dfrac{1}{v}$

The chain rule states that
$ \dfrac{dy}{dx} = \dfrac{du}{dx} \cdot \dfrac{dv}{du} \cdot \dfrac{dy}{dv}$In other words multiply your three derivatives together and simplify

Spoiler

$\dfrac{dy}{dx} = 5 \cdot \cos(5x) \cdot \dfrac{1}{\sin(u)}$
$\dfrac{dy}{dx} = 5 \cdot \cos(5x) \cdot \dfrac{1}{\sin(5x)}$

$\dfrac{dy}{dx} = \dfrac{5cos(5x)}{\sin(5x)}$

Use your trig identities to simplify into the desired format

 

[DeusAbscondus]

Supplementary question: Re: derivative of ln(sin5x)

Super! simply super, thanks for such a full demonstration: really appreciated.
C'n Black: I'm coming to enjoy your laconic exactitude: thank you kindly.
Sudharaka: prompt, friendly, accurate and complete: thanks friend.

It was obvious to me that I was missing something basic: to wit: $$sin(5x)$$calls for the chain rule, since it contains an embedded function.

Supplementary Question: does this mean that whenever the argument of a function - the $x$ bit - contains more than $x$, that one is dealing with an embedded function? (excluding of course the cases of simple derivative addition and subtraction)? Please, could one of you chaps come up with language that tidies up what I think I am right in trying to say here?

this has opened the afternoon up for me to pleasantly work away at 40 examples of a similar kind, so as to root this shared logic into my mathematically growing but immature brain.

DeusAbs

 

[Sudharaka]

Re: Supplementary question: Re: derivative of ln(sin5x)

Super! simply super, thanks for such a full demonstration: really appreciated.
C'n Black: I'm coming to enjoy your laconic exactitude: thank you kindly.
Sudharaka: prompt, friendly, accurate and complete: thanks friend.

You are welcome. :)

Supplementary Question: does this mean that whenever the argument of a function - the $x$ bit - contains more than $x$, that one is dealing with an embedded function? (excluding of course the cases of simple derivative addition and subtraction)? Please, could one of you chaps come up with language that tidies up what I think I am right in trying to say here?

Yes. In mathematical language what you have is a composition of functions. Examples of using the chain rule,

\[f(x)=\sin(5x)\Rightarrow\frac{d}{dx}f(x)=\frac{d}{d(5x)}\sin(5x)\frac{d}{dx}(5x)=5\cos(5x)\]

\[f(x)=\sin(x+10)\Rightarrow \frac{d}{d(x+10)}\sin(x+10)\frac{d}{dx}(x+10)=\cos(x+10)\]

A useful video giving examples of using the chain rule can be found http://www.khanacademy.org/math/calculus/v/chain-rule-examples.

Kind Regards,
Sudharaka.

 

[DeusAbscondus]

Re: Supplementary question: Re: derivative of ln(sin5x)

You are welcome. :)
Yes. In mathematical language what you have is a composition of functions. Examples of using the chain rule,

\[f(x)=\sin(10x)\Rightarrow\frac{d}{dx}f(x)=\frac{d}{d(5x)}\sin(5x)\frac{d}{dx}(5x)=5\cos(5x)\]

\[f(x)=\sin(x+10)\Rightarrow \frac{d}{d(x+10)}\sin(x+10)\frac{d}{dx}(x+10)=\cos(x+10)\]

A useful video giving examples of using the chain rule can be found http://www.khanacademy.org/math/calculus/v/chain-rule-examples.

Kind Regards,
Sudharaka.

Sudharaka, thanks, but are you sure you have not made an inadvertant error here:
$$f(x)=\sin(10x)\implies f'(x)=10cos(10x)$$
doesn't it?
DeusAbs

 

[Sudharaka]

Re: Supplementary question: Re: derivative of ln(sin5x)

Sudharaka, thanks, but are you sure you have not made an inadvertant error here:
$$f(x)=\sin(10x)\implies f'(x)=10cos(10x)$$
doesn't it?
DeusAbs

Yes, sorry. Corrected it. :)

 

[DeusAbscondus]

follow up regarding "special result" in my calc text

I'm trying to satisfy my mind as to why $$ f(x)=5ln(x) \Rightarrow f'(x)=5.\frac{1}{x}=\frac{5}{x}$$

Is it valid because of the result
$$ kf(x) \Rightarrow \frac{dy}{dx}=kf'(x) where 'k' is some constant$$
If so, could someone please make some brief comments illuminating this result, its proof and application (with couple of simple examples)

Thanks kindly folks, always, for the incalculable power of good you are doing me with your help.
DeusAbs

Incidentally,
i) how do I space English words in latex expressions to avoid the ugly running together of letters as in above, and
ii) is there a backslash command for If/then?, or
iii) is my use of $\Rightarrow$ appropriate for If/then expressions?

 

[Jameson]

Hi DeusAbscondus,

Yes, you are correct that for constant k, [math]\frac{d}{dx}k \cdot f(x)=k \cdot f'(x)[/math] and for that reason if [math]f(x)=5\ln(x)[/math] then [math]f'(x)=5 \cdot \frac{1}{x}=\frac{5}{x}[/math].

To answer one of your other questions. You can use the Latex command \text{ } to add normal looking text.

[math]\text{You can see this in effect right here}[/math]

 

[DeusAbscondus]

Hi DeusAbscondus,

Yes, you are correct that for constant k, [math]\frac{d}{dx}k \cdot f(x)=k \cdot f'(x)[/math] and for that reason if [math]f(x)=5\ln(x)[/math] then [math]f'(x)=5 \cdot \frac{1}{x}=\frac{5}{x}[/math].

To answer one of your other questions. You can use the Latex command \text{ } to add normal looking text.

[math]\text{You can see this in effect right here}[/math]

Thanks kindly Jameson (btw: I thoroughly approve of the change of image, which is in keeping with the wry, comic -gently self-ironizing- tone of the former, KGB avatar; i like it!)
I love this place! The sense of support is enormous.
Yesterday I had a day during which I felt defeated by the seeming enormity of the task of learning all this stuff; after a couple of interactions here, I *always* get enlightened and/or instructed, and this, in turn, lifts my mood and encourages me to keep going.

So, now I can go $$\text{If } y=x^2\text{ then } y'=2x\text{...how cool is that}$$

Argggh, but I still have the spacing problem, even after using \text{*} (?!)

 

[Jameson]

I'm really glad you like the site. Hearing that is why I love working here, as well as getting to interact with some great minds.

I've edited your post for you to include the spaces. If you add a space inside the \text{ } then it will appear. For example:

1) \text{The derivative of} x^2 \text{is} 2x yields [math]\text{The derivative of} x^2 \text{is} 2x[/math]. Notice the problem in spacing.

2) If I add spaces inside the text tag like so: \text{The derivative of } x^2 \text{ is } 2x we get [math]\text{The derivative of } x^2 \text{ is } 2x[/math]

See the difference? :)

 

[Sudharaka]

Re: follow up regarding "special result" in my calc text

Hi DeusAbscondus, :)

I'm trying to satisfy my mind as to why $$ f(x)=5ln(x) \Rightarrow f'(x)=5.\frac{1}{x}=\frac{5}{x}$$

Is it valid because of the result
$$ kf(x) \Rightarrow \frac{dy}{dx}=kf'(x) where 'k' is some constant$$
If so, could someone please make some brief comments illuminating this result, its proof and application (with couple of simple examples)

This is a consequence of the constant factor rule in differentiation. You can find the proof on this page. Examples can be found here.

iii) is my use of $\Rightarrow$ appropriate for If/then expressions?

The symbol \(\Rightarrow\) is used for implication. Since \(y=kf(x)\) implies \(\frac{dy}{dx}=kf'(x)\) it is correct to write, \(y=kf(x)\Rightarrow \frac{dy}{dx}=kf'(x)\). Refer this.

Kind Regards,
Sudharaka.

 

[CaptainBlack]

Re: follow up regarding "special result" in my calc text

I'm trying to satisfy my mind as to why $$ f(x)=5ln(x) \Rightarrow f'(x)=5.\frac{1}{x}=\frac{5}{x}$$

Is it valid because of the result
$$ kf(x) \Rightarrow \frac{dy}{dx}=kf'(x) where 'k' is some constant$$
If so, could someone please make some brief comments illuminating this result, its proof and application (with couple of simple examples)

Thanks kindly folks, always, for the incalculable power of good you are doing me with your help.
DeusAbs

Incidentally,
i) how do I space English words in latex expressions to avoid the ugly running together of letters as in above, and
ii) is there a backslash command for If/then?, or
iii) is my use of $\Rightarrow$ appropriate for If/then expressions?

By definition:

\[\frac{d}{dx}f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\]

so:

\[\begin{aligned} \frac{d}{dx}\left(k\; f(x)\right)&=\lim_{h \to 0}\frac{k\;f(x+h)-k\;f(x)}{h} \\ &= \lim_{h \to 0}\left(k\; \frac{f(x+h)-f(x)}{h}\right)\\ &= k\; \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &= k\frac{d}{dx}f(x) \end{aligned}\]

CB

 

[DeusAbscondus]

Re: follow up regarding "special result" in my calc text

By definition:

\[\frac{d}{dx}f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\]

so:

\[\frac{d}{dx}\left(k\; f(x)\right)=\lim_{h \to 0}\frac{k\;f(x+h)-k\;f(x)}{h} = \lim_{h \to 0}\left(k\; \frac{f(x+h)-f(x)}{h}\right) = k\; \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\ = k\frac{d}{dx}f(x)\]

CB

Thanks Cap'n.
Could you go one step further and instance the use of $\frac{dy}{dx}$ as opposed to $\frac{d}{dx}$ so as to make plain to me the logic and syntax of the one compared to the other?
Much appreciated,
DeusAbs

 

[Jameson]

Re: follow up regarding "special result" in my calc text

Thanks Cap'n.
Could you go one step further and instance the use of $\frac{dy}{dx}$ as opposed to $\frac{d}{dx}$ so as to make plain to me the logic and syntax of the one compared to the other?
Much appreciated,
DeusAbs

Usually [math]\frac{dy}{dx}=f'(x)=y'[/math] meaning the derivative of function y with respect to x. Nothing necessarily needs to be calculated. All three of those things just represent the idea of the derivative of y with respect to x.

On the other hand, [math]\frac{d}{dx}[/math] means calculate the derivative of whatever follows with respect to x.

Example:
[math]y=x^2[/math] so [math]\frac{d}{dx}x^2=2x=\frac{dy}{dx}[/math]

With [math]\frac{d}{dx}[/math] something must follow that or it doesn't make sense. [math]\frac{dy}{dx}[/math] can stand on its own.

 

[DeusAbscondus]

I'm really glad you like the site. Hearing that is why I love working here, as well as getting to interact with some great minds.

I've edited your post for you to include the spaces. If you add a space inside the \text{ } then it will appear. For example:

1) \text{The derivative of} x^2 \text{is} 2x yields [math]\text{The derivative of} x^2 \text{is} 2x[/math]. Notice the problem in spacing.

2) If I add spaces inside the text tag like so: \text{The derivative of } x^2 \text{ is } 2x we get [math]\text{The derivative of } x^2 \text{ is } 2x[/math]

See the difference? :)

Ah! Formidable, mon vieux!
This is exactly what I came on to find out about tonight! and here, you left me word to that effect days ago!
$\text{ Thanks again ... Jameson ... }$:)