Differentiation Question (1 Viewer)

[dragon658]

Hi guys can someone explain how to differentiate sin(3x+1)

I've been through the worked solutions and I guess it makes sense but it doesn't make sense in regard to the Chain Rule I've been taught - [f(x)]^n = n[f(x)]^n-1 x f'(x)

Thanks.

 

[InteGrand]

Hi guys can someone explain how to differentiate sin(3x+1)

I've been through the worked solutions and I guess it makes sense but it doesn't make sense in regard to the Chain Rule I've been taught - [f(x)]^n = n[f(x)]^n-1 x f'(x)

Thanks.

The form of the chain rule you've quoted is for when our function is raised to some power, which isn't the case here. The form of the chain rule we'll want for this is:

d/dx (f(ax+b)) = a*f'(ax+b).

So here, d/dx (sin(3x+1)) = 3*cos(3x+1).

 

[InteGrand]

You can use the chain rule explicitly as follows.

Let y = sin(3x+1). Let u = 3x+1. So y = sin u.

Now, dy/du = cos u, and du/dx = 3. So

dy/dx = (dy/du)*(du/dx)

= (cos u)*3

= 3*cos u

= 3*cos(3x+1).

With enough practice, you will probably be able to write down the answers to things like this straightaway.

 

[dragon658]

Ok I see. I went through it this morning and it made more sense. I think I thought I had to differentiate sin(3x+1) as a whole instead of sin and 3x+1 as separate things. Also, I just watched a video on HSCHub about this and I didn't even really there was another kind of, form of the Chain rule.
Thanks again.

 

[Yagami Light]

With enough practice, you will probably be able to write down the answers to things like this straightaway.

but if the question asks you to use chain rule do you have to write down the working out like dy/dx=dy/du*du/dx or can you just write the answer straightaway?

 

[si2136]

but if the question asks you to use chain rule do you have to write down the working out like dy/dx=dy/du*du/dx or can you just write the answer straightaway?

Pretty sure you're allowed to write it straight away, because it's by inspection.

Eg. (By Inspection) = Answer

 

[LC14199]

How I was taught to do these is simply just take the derivative of what's inside the brackets, put it outside, and then know what the derivative of the trig function is. eg Sin -> Cos, Cos -> -Sin and so on.

 

[si2136]

How I was taught to do these is simply just take the derivative of what's inside the brackets, put it outside, and then know what the derivative of the trig function is. eg Sin -> Cos, Cos -> -Sin and so on.

That's how you do it lol

 

[LC14199]

That's how you do it lol

Well, op asked for an explanation, so he got one xD Sometimes worded explanations can make more sense than hundreds of mathematical examples lol.

 

[kashkow]

but if the question asks you to use chain rule do you have to write down the working out like dy/dx=dy/du*du/dx or can you just write the answer straightaway?

Pretty sure you can write it straight away but I'd ask your teacher (you may need to write the formula in certain cases, like if you are learning it).

How I was taught to do these is simply just take the derivative of what's inside the brackets, put it outside, and then know what the derivative of the trig function is. eg Sin -> Cos, Cos -> -Sin and so on.

That is a good way to think of it and OP I would recommend learning this word way of learning it to help you know when/how to apply (basically same as what you said, LC, but other way around):

If there's a function inside a function, find the derivative of the outer function (keeping the inside function the same) and multiply the derivative of the inside function.

So like for the above example in the quote, d/dx[sin(3x+1)] = cos(3x+1) * 3.

I've bolded and italicised to help see what part is what. It then becomes identifying when there is a function of a function and which is outer function/inner function. This works for pretty much everything and you can apply this multiple times if there is a function inside a function inside a function...

 

[LC14199]

Pretty sure you can write it straight away but I'd ask your teacher (you may need to write the formula in certain cases, like if you are learning it).



That is a good way to think of it and OP I would recommend learning this word way of learning it to help you know when/how to apply (basically same as what you said, LC, but other way around):

If there's a function inside a function, find the derivative of the outer function (keeping the inside function the same) and multiply the derivative of the inside function.

So like for the above example in the quote, d/dx[sin(3x+1)] = cos(3x+1) * 3.

I've bolded and italicised to help see what part is what. It then becomes identifying when there is a function of a function and which is outer function/inner function. This works for pretty much everything and you can apply this multiple times if there is a function inside a function inside a function...

Duuuude use LaTeX code lmao. It would've been in there from when you quoted me xD Makes it so much prettier xD

 

[kashkow]

Duuuude use LaTeX code lmao. It would've been in there from when you quoted me xD Makes it so much prettier xD

LOL I don't know how xD well true, I could've copied it from you, but it looked so confusing and it doesn't really need it as it's clear enough (just basic sins, and equations, no powers) (plus it is/was a little different to what you wrote - originally i didn't have d/dx and the order of the terms in the derivative is slightly different).

If I can do Latex with bold and italicised text I will do it (if I can figure it out with the text I wrote). Otherwise I'll leave it for clarity.



EDIT: ahahaha I can't figure out how to do bold and italics with Latex, so im just going to leave it .

 

[LC14199]

Odd, \boldmath doesn't work on the forums. I will investigate this matter further.

Edit: It appears the forums is using some strange version of LaTeX... give me some time lol.

Edit 2: \pmb{} works, but it makes it UBER bold and ugly. Let me try and find a working normal bold font on this forum before resorting to that.

Edit 3: I've updated my old post with another version underneath, but the x does not have the nice curves on it when bolded, something that frustrates me immensely. I'm looking for a fix now.

Edit 4: Fu**ing finally. Got it. Posting the nice version here



This is the LaTeX code for the above statement: \frac{d}{dx}[\sin(3x+1)]=\mathbf{3\boldsymbol{\cos}(3\textbf{\textit{{x}}}\boldsymbol{+}1)}

This is the LaTeX code for the below statement:
\frac{d}{dx}[\sin(3x+1)]=\mathbf{{\boldsymbol{\cos}(3\textbf{\textit{{x}}}\boldsymbol{+}1)\times3}}





Now, having succeeded in that, I am going to bed as it's 2:30 in the morning. Goodnight gentlemen. Enjoy your bolded, pretty equation