Trigonometric Derivative Question

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Jason76

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What is the reasoning behind:

----------

Given:

derivative of sin x = cos x

\(\displaystyle sin x = cos x\)

derivative of

\(\displaystyle cos x = - sin x\)

derivative of

\(\displaystyle tan x = sec^{2}x\)

------------

\(\displaystyle y = sin (x + b)\)

answer:

\(\displaystyle cos (x + b)\)

\(\displaystyle y = sin ax^{2}\)

answer:

\(\displaystyle 2 ax cos ax^{2}\)

\(\displaystyle y = tan(3\) theta)

answer:

\(\displaystyle 3 sec^{2} (3\) theta)
 
Last edited:
It's mathematics. We're not just making stuff up. Please consult the definition of the derivative.
 
Ok, let's look at the first one:

\(\displaystyle y = sin (x + b)\)

rewrite in substitution terms:

\(\displaystyle y = sin (u)\)

Find derivatives of y and u

\(\displaystyle Dy = cos (u)\)

\(\displaystyle Du = 1 \)

Final answer is

\(\displaystyle Dy(Du)\)

so

\(\displaystyle y = [Cos(x + b)](1)\) Final Answer

But why was the derivative of u 1?

-----

Looking at problem 2

\(\displaystyle y = sinax^{2}\)

rewritten as

\(\displaystyle y = sin (u)\)

a can be interpreted as a number substitute.

so

\(\displaystyle u = ax^{2}\)

\(\displaystyle Du = 2(a)(x)\)

or

\(\displaystyle Du = 2ax\)

now

\(\displaystyle Dy = cos(u)\)

So now plugging in numbers:

\(\displaystyle y = (cos ax^{2}) (2ax)\) is the final answer.

because

\(\displaystyle y = Dy(Du)\)
 
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Jason76 said:
What is the reasoning behind:
derivative of sin x = cos x
Click to expand...
Your book does not demonstrate?

\(\displaystyle y = sin(x).\)

\(\displaystyle y + \Delta y = sin(x + \Delta x) = \{sin(x) * cos(\Delta x)\} + \{cos(x) * sin(\Delta x)\}.\) Basic trigonometry.

\(\displaystyle \Delta y = sin(x + \Delta x) - sin(x) = \{sin(x) * cos(\Delta x)\} + \{cos(x) * sin(\Delta x)\} - sin(x) = sin(x)\{cos(\Delta x) - 1\} + \{cos(x) * sin(\Delta x)\}.\)

\(\displaystyle \dfrac{\Delta y}{\Delta x} = \left\{sin(x) * \dfrac{cos(\Delta x) - 1}{\Delta x}\right\} + \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\} = \left\{sin(x) * \dfrac{cos(\Delta x) - 1}{\Delta x} * \dfrac{cos(\Delta x) + 1}{cos(\Delta x) + 1}\right\} + \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\} =\)

\(\displaystyle \left\{sin(x) * \dfrac{- sin^2(\Delta x)}{\Delta x[cos(\Delta x) + 1]}\right\} + \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\} = \left\{- \dfrac{sin(x) * sin(\Delta x)}{cos(\Delta x) + 1} * \dfrac{sin(\Delta x)}{\Delta x}\right\}+ \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\}.\)

\(\displaystyle But\ \displaystyle \lim_{\Delta x \rightarrow 0}\left(\dfrac{sin(\Delta x)}{\Delta x}\right) = 1.\)

\(\displaystyle And\ \displaystyle \lim_{\Delta x \rightarrow 0}\left(- \dfrac{sin(x) * sin(\Delta x)}{cos(\Delta x) + 1}\right) = -sin(x) * \lim_{\Delta x \rightarrow 0}\dfrac{sin(\Delta x)}{cos(\Delta x) + 1} = -sin(x) * \dfrac{\displaystyle \lim_{\Delta x \rightarrow 0}sin(\Delta x)}{\displaystyle \lim_{\Delta x \rightarrow 0}cos(\Delta x) + 1} = -sin(x) * \dfrac{0}{2} = 0.\)

\(\displaystyle So\ \displaystyle \lim_{\Delta x \rightarrow 0}\dfrac{\Delta y}{\Delta x} = \dfrac{dy}{dx} = \{0 * 1\} + \{cos(x) * 1\} = cos(x).\)

You can do the rest by using trig identities and basic propositions about derivatives.

\(\displaystyle y = cos(x) = \sqrt{1 - sin^2(x)}.\)

\(\displaystyle Let\ u = sin(x) \implies \dfrac{du}{dx} = cos(x).\)

\(\displaystyle Let\ v = 1 - u^2 \implies \dfrac{dv}{du} = - 2u.\)

\(\displaystyle So\ y = \sqrt{v} \implies \dfrac{dy}{dv} = \dfrac{1}{2\sqrt{v}} \implies\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{dv} * \dfrac{dv}{du} * \dfrac{du}{dx} = \dfrac{1}{2\sqrt{v}} * (-2u) * cos(x) = \dfrac{-2sin(x) * cos(x)}{2\sqrt{1 - sin^2(x)}} = - \dfrac{sin(x) * cos(x)}{cos(x)} = -sin(x).\)
 
JeffM said:
Your book does not demonstrate?

\(\displaystyle y = sin(x).\)

\(\displaystyle y + \Delta y = sin(x + \Delta x) = \{sin(x) * cos(\Delta x)\} + \{cos(x) * sin(\Delta x)\}.\) Basic trigonometry.

\(\displaystyle \Delta y = sin(x + \Delta x) - sin(x) = \{sin(x) * cos(\Delta x)\} + \{cos(x) * sin(\Delta x)\} - sin(x) = sin(x)\{cos(\Delta x) - 1\} + \{cos(x) * sin(\Delta x)\}.\)

\(\displaystyle \dfrac{\Delta y}{\Delta x} = \left\{sin(x) * \dfrac{cos(\Delta x) - 1}{\Delta x}\right\} + \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\} = \left\{sin(x) * \dfrac{cos(\Delta x) - 1}{\Delta x} * \dfrac{cos(\Delta x) + 1}{cos(\Delta x) + 1}\right\} + \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\} =\)

\(\displaystyle \left\{sin(x) * \dfrac{- sin^2(\Delta x)}{\Delta x[cos(\Delta x) + 1]}\right\} + \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\} = \left\{- \dfrac{sin(x) * sin(\Delta x)}{cos(\Delta x) + 1} * \dfrac{sin(\Delta x)}{\Delta x}\right\}+ \left\{cos(x) * \dfrac{sin(\Delta x)}{\Delta x}\right\}.\)

\(\displaystyle But\ \displaystyle \lim_{\Delta x \rightarrow 0}\left(\dfrac{sin(\Delta x)}{\Delta x}\right) = 1.\)

\(\displaystyle And\ \displaystyle \lim_{\Delta x \rightarrow 0}\left(- \dfrac{sin(x) * sin(\Delta x)}{cos(\Delta x) + 1}\right) = -sin(x) * \lim_{\Delta x \rightarrow 0}\dfrac{sin(\Delta x)}{cos(\Delta x) + 1} = -sin(x) * \dfrac{\displaystyle \lim_{\Delta x \rightarrow 0}sin(\Delta x)}{\displaystyle \lim_{\Delta x \rightarrow 0}cos(\Delta x) + 1} = -sin(x) * \dfrac{0}{2} = 0.\)

\(\displaystyle So\ \displaystyle \lim_{\Delta x \rightarrow 0}\dfrac{\Delta y}{\Delta x} = \dfrac{dy}{dx} = \{0 * 1\} + \{cos(x) * 1\} = cos(x).\)

You can do the rest by using trig identities and basic propositions about derivatives.

\(\displaystyle y = cos(x) = \sqrt{1 - sin^2(x)}.\)

\(\displaystyle Let\ u = sin(x) \implies \dfrac{du}{dx} = cos(x).\)

\(\displaystyle Let\ v = 1 - u^2 \implies \dfrac{dv}{du} = - 2u.\)

\(\displaystyle So\ y = \sqrt{v} \implies \dfrac{dy}{dv} = \dfrac{1}{2\sqrt{v}} \implies\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{dv} * \dfrac{dv}{du} * \dfrac{du}{dx} = \dfrac{1}{2\sqrt{v}} * (-2u) * cos(x) = \dfrac{-2sin(x) * cos(x)}{2\sqrt{1 - sin^2(x)}} = - \dfrac{sin(x) * cos(x)}{cos(x)} = -sin(x).\)
Click to expand...

Thanks, but I think what I really mean was "How do the answers to the particular problems come about?". But I worked out 2 problems. But how about the

\(\displaystyle x + b\) and it's derivative of 1.
 
Jason76 said:
Thanks, but I think what I really mean was "How do the answers to the particular problems come about?". But I worked out 2 problems. But how about the

\(\displaystyle x + b\) and it's derivative of 1.
Click to expand...
Jason

The derivative of x + b with respect to x is 1. Work it out from first principles if necessary.

But when you assumed that the derivative of sin(u) was cos(u), you were assuming what I thought you were trying to prove, namely that the derivative of sin(x) is cos(x). Such a proof is circular and so invalid.

Perhaps I do not understand what you were asking. If so, I am sorry to have wasted your time.
 
JeffM said:
Jason

The derivative of x + b with respect to x is 1. Work it out from first principles if necessary.

But when you assumed that the derivative of sin(u) was cos(u), you were assuming what I thought you were trying to prove, namely that the derivative of sin(x) is cos(x). Such a proof is circular and so invalid.

Perhaps I do not understand what you were asking. If so, I am sorry to have wasted your time.
Click to expand...

No, nobody is wasting my time. Everything learned is useful.
 
Jason76 said:
What is the reasoning behind:

----------

Given:

derivative of sin x = cos x

\(\displaystyle sin x = cos x\) . . . . . False. There is no equality here. Do not use an equals sign.

derivative of

\(\displaystyle cos x = - sin x\)

derivative of

\(\displaystyle tan x = sec^{2}x\)

------------

\(\displaystyle y = sin (x + b)\)

answer:

y' = \(\displaystyle cos (x + b)\)


\(\displaystyle y \ = \ sin(ax^{2})\) . . . . . Use grouping symbols, such as parentheses.

answer:

y' = \(\displaystyle 2axcos(ax^{2})\) . . . . .Use grouping symbols, such as parentheses.


\(\displaystyle y = tan(3\theta)\)

answer:

y' = \(\displaystyle 3sec^{2} (3\theta)\)
Click to expand...

Jason76,

you are inconsistent (and relatively careless) with your notation.


Let's use mostly English first and then mostly mathematical symbols for the following example.


The derivative of sin(x) is cos(x).


\(\displaystyle \dfrac{d}{dx}[sin(x)] \ = \ cos(x).\)



\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ y \ = \ sin(x)\)


Then \(\displaystyle \ \ y' \ = \ cos(x)\)
 
lookagain said:
Jason76,

you are inconsistent (and relatively careless) with your notation.


Let's use mostly English first and then mostly mathematical symbols for the following example.


The derivative of sin(x) is cos(x).


\(\displaystyle \dfrac{d}{dx}[sin(x)] \ = \ cos(x).\)



\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ y \ = \ sin(x)\)


Then \(\displaystyle \ \ y' \ = \ cos(x)\)
Click to expand...

Sorry, working on improving my La-Tex notation. I will try to make my English explanations clearer, as well as use all La-Tex symbols for easier reading.
 
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The "reasoning" is the chain rule.

If [FONT=MathJax_Math-italic-Web]y[/FONT][FONT=MathJax_Main-Web]=[/FONT][FONT=MathJax_Math-italic-Web]s[/FONT][FONT=MathJax_Math-italic-Web]i[/FONT][FONT=MathJax_Math-italic-Web]n[/FONT][FONT=MathJax_Main-Web]([/FONT][FONT=MathJax_Math-italic-Web]x[/FONT][FONT=MathJax_Main-Web]+[/FONT][FONT=MathJax_Math-italic-Web]b[/FONT][FONT=MathJax_Main-Web])[/FONT] y= sin(x+ b) then the derivative of y is cos(x+ b) times the derivative of x+ b, which is 1. y'= cos(x+ b).

If \(\displaystyle y= cos(ax^2)\) then the derivative of y is \(\displaystyle -sin(ax^2)\) times the derivative of \(\displaystyle ax^2\), which is 2ax: \(\displaystyle y'= -2ax sin(ax^2)\).

If \(\displaystyle y= tan(3\theta)\) then the derivative of y (with respect to \(\displaystyle \theta\) is \(\displaystyle sec^2(3\theta)\) times the derivative of \(\displaystyle 3\theta\), which is 3: \(\displaystyle y'= 3sec^2(3\theta)\).
 
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