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Introduction / Summary of Differentiation

  1. Oct 23, 2006 #1


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    As the title suggests this thread is intended as a summary of differentiation, it is by no means an attempt at an exhaustive discussion of the topic. A detailed knowledge of limits is not required but is useful, it is assumed that the reader has some knowledge of limits and how to implement them. As always I welcome comments and corrections either here of via PM.

    The Definition of a Derivative

    The derivative ([itex]f'(x)[/itex]) of a function [itex]f(x)[/itex] with respect to [itex]x[/itex] is given by the limit;


    Provided the limit exists. If the limit exist at a point x0 we say that the function is differentiable at this point.

    As can be seen from the plots below, as [itex]h[/itex] approaches zero, the secants of the curve tend to the tangent at a point x, this gives the gradient of the curve at point x (provided the function is differentiable at point x). Note however, that in the above limit we cannot simply set [itex]h[/itex] as zero directly as our limit would [itex]\to\pm\infty[/itex] and thus, the limit would not exist. We must therefore manipulate the limit into a form where this does not occur.

    [​IMG] [​IMG]
    Images taken from Wikipedia

    Find the derivative of [itex]f(x)= x^2+2[/itex]
    [tex]f'(x)=\lim_{h\to0}\;\frac{f(x+h)-f(x)}{h}\;\;\rightarrow \lim_{h\to0}\;\frac{(x+h)^2+2-(x^2+2)}{h}[/tex]
    [tex]=\lim_{h\to0}\;\frac{\not{x^2}+h^2+2hx +\not{2}-\not{x^2}-\not{2}}{h}= lim_{h\to0}\;\frac{\not{h}(h+2x)}{\not{h}}[/tex]
    [tex]\therefore f'(x)=2x[/tex]​

    Finding the Gradient and Tangent to a Curve at a Point
    We already now that the derivative of a function gives the gradient of a curve at any point that is differentiable. This allows us to find the gradient of the function at any point x where the curve is differentiable. This then allows us to find the equation of the tangent to the curve at any point x where the function is differentiable.

    Find the equation of the tangent to the curve with equation [itex]y=f(x)=x^2+2[/itex] at x=3.
    We know that the tangent is a straight line and therefore has an equation of the form [itex]y=mx+c[/itex]. We have already found the derivative above which gives the gradient (m) of any point x on the curve y=f(x) (provided f(x) is differentiable at x). Therefore we can say that;

    [tex]m=\left. \frac{dy}{dx}\;\;\right|_{x=3}=2\cdot3=6[/tex]

    We can find the corresponding y value to x=3 using the original equation [itex]y=x^2+2[/itex];
    We can then use this information to find the equation of the tangent at point [itex](x,y)=(3,11)[/itex];
    [tex]y=mx+c \rightarrow 11=3\cdot 6 + C \Rightarrow C = -7[/tex]
    [itex]\therefore[/itex] the equation of the tangent to the curve [itex]y=x^2+2[/itex] at the point [itex](x,y)=(3,11)[/itex] is [itex]y=6x-7[/itex].​
    Last edited: Oct 23, 2006
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  3. Oct 23, 2006 #2


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    Rules of Differentiation

    Using the definition of a derivative every time we wish to differentiate a function becomes tedious after a while. It is therefore desirable to derive a series of rules which allows us to differentiate a variety of functions without having to refer to the above definition.

    1. Derivative of a Constant Function
    If [itex]f[/itex] has a constant value [itex]f(x)=c[/itex], then;


    [tex]\therefore {\color{red}\boxed{{\color{black}\frac{d}{dx}\;(c) = 0}}}[/tex]

    2. Power Rule for Positive Integers
    If [itex]f(x)=x^{n}[/itex] and [itex] n\in\mathbb{Z},\;n>0[/itex] then;

    Since n is a positive integer we can use the binomial theorem to expand the parentheses;

    [tex]=\lim_{h\to0}\frac{\left\{\not{x^n} + nx^{-1}\cdot h + \frac{n(n-1)}{2!}\cdot x^{n-2}h^{2} + \ldots + nxh^{n-1}+h^{n}\right\}-\not{x^{n}}}{h}[/tex]

    [tex]=\lim_{h\to0}\frac{nx^{-1}\cdot h + \frac{n(n-1)}{2!}\cdot x^{n-2}h^{2} + \ldots + nxh^{n-1}+h^{n}}{h}[/tex]

    One h from each term in the numerator will cancel with the h is the denominator leaving;

    [tex]=\lim_{h\to0}\; nx^{-1} + \frac{n(n-1)}{2!}\cdot x^{n-2}h + \ldots + nxh^{n-2}+h^{n-1}[/tex]

    As [itex]h\to0[/itex] all terms except the first drop out leaving;


    [tex]\therefore {\color{red}\boxed{{\color{black}\frac{d}{dx}\;n^{n} = nx^{n-1}\;\;\; n\in\mathbb{Z},\;n>0}}}[/tex]

    Note that this can be shown to be true for all integers (which will be done later).

    3. Constant Coefficient Rule
    If [itex]f[/itex] is a differentiable function of [itex]x[/itex] and [itex]c[/itex] is constant then;

    [tex]f'(x)=\lim_{h\to0}\frac{cf(x+h)-cf(x)}{h} \stackrel{Limit\;Property}{=} c\cdot\lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]

    Since the RHS of the equality is the definition of the derivative we can say that;

    [tex]{\color{red}\boxed{{\color{black}\frac{d}{dx}\;(cf(x)) = c\cdot\frac{df}{dx}}}}[/tex]

    4. The Derivative of a Sum
    If we take two functions of [itex]x[/itex]; [itex]u[/itex] and [itex]v[/itex], both of which are differentiable then;

    [tex]\frac{d}{dx} \left(u(x)+v(x)\right) = \lim_{h\to0}\frac{(u(x+h)+v(x+h))-(u(x)+v(x))}{h}[/tex]

    We can then split this into two fractions;

    [tex]=\lim_{h\to0}\left\{ \frac{u(x+h)-u(x)}{h}+\frac{v(x+h)-v(x)}{h}\right\}[/tex]

    [tex]=\lim_{h\to0}\frac{u(x+h)-u(x)}{h} + \lim_{h\to0}\frac{v(x+h)-v(x)}{h}\right\} = \frac{du}{dx}+\frac{dv}{dx}[/tex]

    [tex]\therefore {\color{red}\boxed{{\color{black}\frac{d}{dx}\;(u+v)=\frac{du}{dx}+\frac{dv}{dx}}}}[/tex]

    The sum rule can be shown to hold for differences (i.e. [itex](u-v)[/itex]) by combining the sum rule with the constant multiple rule and setting [itex]c=-1[/itex].
    Last edited: Oct 23, 2006
  4. Oct 28, 2006 #3


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    5. Product Rule
    Suppose we have two functions of [itex]x[/itex] - [itex]u(x)[/itex] and [itex]v(x)[/itex] both differentiable at [itex]x[/itex] - and we wish to find the derivative of their product, then from our definition of the derivative we can write;

    [tex]\frac{d}{dx}\;(u\dot v) = \lim_{h\to0}\frac{u(x+h)\cdot v(x+h)-u(x)\cdot v(x)}{h}[/tex]

    The next step may appear nonsensical initially but is necessary so we can evaluate the limits of [itex]u[/itex] and [itex]v[/itex] separately. The next step is to add and then subtract [itex]u(x+h)\cdot v(x)[/itex] from the numerator (the additions are highlighted in blue);

    [tex]=\lim_{h\to0}\frac{u(x+h)\cdot v(x+h) {\color{blue}-u(x+h)\cdot v(x) +u(x+h)\cdot v(x)} -u(x)\cdot v(x)}{h}[/tex]
    [tex]=\lim_{h\to0}\left\{ \frac{u(x+h)\cdot v(x+h) {\color{blue}-u(x+h)\cdot v(x)}}{h}+\frac{{\color{blue}u(x+h)\cdot v(x)} - u(x)\cdot v(x)}{h}\right\}[/tex]

    We can now take a factor of [itex]u(x+h)[/itex] from the first ratio and a factor of [itex]v(x)[/itex] from the second ratio, giving;

    [tex]=\lim_{h\to0}\left\{ u(x+h)\cdot\frac{v(x+h)-v(x)}{h}+v(x)\cdot\frac{u(x+h)-u(x)}{h}\right\} = \lim_{h\to0}\; u(x+h)\cdot \lim_{h\to0}\frac{v(x+h)-v(x)}{h}+\lim_{h\to0}\frac{u(x+h)-u(x)}{h}[/tex]

    Since we stated that both [itex]u[/itex] and [itex]v[/itex] are differentiable at [itex]x[/itex] this implies that both are continuous* at [itex]x[/itex], therefore as [itex]h[/itex] approaches zero, [itex]u(x+h)[/itex] approaches [itex]u(x)[/itex]. Therefore, the above limits become what is known as the product rule;

    [tex]{\color{red}\boxed{{\color{black}\frac{d}{dx}\;(u\cdot v)= u\cdot\frac{dv}{dx}+v\cdot\frac{du}{dx}}}}[/tex]

    [tex]\frac{d}{dx}\left\{ (x^4+4)\cdot (x^2-1)\right\}[/tex]

    Let [itex]u(x):=x^4+4[/itex] and [itex]v(x):=x^2-1[/itex]. First we must find [itex]u'(x)[/itex] and [itex]v'(x)[/itex] for this we can use the power and the constant rule (I will not show all the steps here since the application of the rules are simple) which gives;

    [tex]\frac{du}{dx} = 4x^3 \;\;\;\;\;\;\;\;\ \frac{dv}{dx} = 2x[/tex]

    Therefore, using the product rule we obtain;

    [tex]\frac{d}{dx} (x^4+4)\cdot (x^2-1) =u\frac{dv}{dx}+v\frac{du}{dx}= (x^4+4)\cdot 2x + (x^2-1)\cdot 4x^2 = 2(x^5 + 2x^4 -2x^2 + 4x)[/tex]

    *To prove this would require a discussion of limits, which I have already said I will not do here. If you wish to see a proof of this, then I will be happy to oblige via PM but a knowledge of limits is required beforehand.
    Last edited: Oct 29, 2006
  5. Oct 29, 2006 #4


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    6. The Quotient Rule
    Suppose we again have two functions of [itex]x[/itex] ([itex]u(x)[/itex] and [itex]v(x)[/itex]), both of which are differentiable at [itex]x[/itex] and [itex]v(x)\neq0[/itex]. What if we wish to find the derivative of the ratio of the two functions?

    [tex]\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \lim_{h\to0}\frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}[/tex]

    [tex]=\lim_{h\to0}\frac{v(x)\cdot u(x+h) - u(x)\cdot v(x+h)}{h\cdot v(x+h)\cdot v(x)}[/tex]

    Again, here we must take the seemingly nonsensical step of adding and subtracting [itex]u(x)\cdot v(x)[/itex] from the numerator. However, this will allow us to write two difference ratios for the derivatives of [itex]u(x)[/itex] and [itex]v(x)[/itex] (again this step is highlighted in blue);

    [tex]=\lim_{h\to0}\frac{v(x)\cdot u(x+h) {\color{blue}-u(x)\cdot v(x) + u(x)\cdot v(x)} -u(x)\cdot v(x+h)}{h\cdot v(x+h)\cdot v(x)}[/tex]

    [tex]=\lim_{h\to0}\frac{v(x)\cdot{\color{red}\frac{u(x+h)-u(x)}{h}}-u(x)\cdot{\color{red}\frac{v(x+h)-v(x)}{h}}}{v(x+h)\cdot v(x)}[/tex]

    Notice that the two ratios highlighted in red are the derivative of the functions [itex]u(x)[/itex] and [itex]v(x)[/itex] respectively. Notice also that as [itex]h\to0[/itex], then [itex]v(x+h)\cdot v(x) \to (v(x))^2[/itex]. Therefore when we evaluate our limit we obtain;



    Let [itex]u(x)=x^2+1[/itex] and [itex]v(x)=x^2-1[/itex], then,

    [tex]\frac{du}{dx}=2x \;\;\;\;\;\; \frac{dv}{dx}=2x\;\;\;\;\;\; (v(x))^2 = (x^2-1)^2[/tex]

    [tex]\Rightarrow \frac{d}{dx}\frac{x^2+1}{x^2-1} = \frac{2x\cdot(x^2+1) - 2x\cdot(x^2-1)}{(x^2-1)^2} = \frac{4x^3+4x}{(x^2-1)}[/tex]
    Last edited: Oct 29, 2006
  6. Nov 9, 2006 #5


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    7. Power Rule for Negative Integers
    Suppose we have [itex]x^n[/itex], where [itex]n \in \mathbb{Z},\; n<0[/itex] and [itex]n=-m[/itex], where [itex]m \in \mathbb{Z},\; n>0[/itex]. Hence we can say that;


    Therefore, using the quotient rule (and positive power rule) we can say that;

    [tex]\frac{d}{dx}\;x^n = \frac{d}{dx}\;\frac{1}{x^m} = \frac{x^m\cdot\frac{d}{dx}(1) - 1\cdot\frac{d}{dx}\;x^m}{(x^m)^2}[/tex]

    [tex]=\frac{0-mx^{m-1}}{x^{2m}} = -mx^{-m-1} \stackrel{n=-m}{=}nx^{n-1}[/tex]

    This result is identical to the power rule for positive integers, we can therefore generalise the power rule thus;

    [tex]\therefore {\color{red}\boxed{{\color{black}\frac{d}{dx}\;n^{ n} = nx^{n-1}\;\;\; n\in\mathbb{Z}}}}[/tex]

    I do not think an example is necessary here.
  7. Nov 9, 2006 #6
    Can you continue proving the Power rule for all real numbers using logarithmic differentiation.
  8. Nov 10, 2006 #7


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    Hi Minase,
    Thank you for your comments. I am planning to discuss the definition and differentiation of logarithmic and exponential functions later in the tutorial. Prior to this I will consider the chain rule, implicit differentiation and differentiation of trigonometric functions.
  9. Nov 10, 2006 #8


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    8. The Chain Rule

    Note; This proof requires knowledge of Linear Approximations and Differentials. For those who have not met linearization or differentials before, I will provide a external link to a 'rough proof' later and you may skip the following section.


    The chain rule is used to find the derivative of a composite function (in other words a function of a function).

    Provided [itex]f(x)[/itex] is differentiable at [itex]\delta[/itex], then we can show that the respective change in [itex]f[/itex] as [itex]\delta[/itex] changes to [itex]\delta+\Delta x[/itex] is given by;

    [tex]\underbrace{f(x+\delta)-f(x)}_{\text{Change in }f}=\underbrace{f'(\delta)}_{\text{approximation}}+\underbrace{\epsilon\Delta x}_{\text{error}}[/tex]
    Where [itex]\Delta x \to 0\Rightarrow \epsilon\to0[/itex].

    Let [itex]u=g(x)[/itex] be differentiable at [itex]x[/itex] and [itex]f(u)[/itex] be differentiable at [itex]g(x)[/itex]. Also, let [itex]\Delta x, \Delta y, \Delta u[/itex] be the respective changes in [itex]x, y, x[/itex]. Then from the above equation form the following system;

    [tex]\Delta u = g'(x)\Delta x + \epsilon_{0}\Delta x[/tex]
    [tex]\Delta y = f'(u)\Delta u + \epsilon_{1}\Delta u[/tex]

    Substituting [itex]\Delta u[/itex] into [itex]\Delta y[/itex] gives;

    [tex]\Delta y = f'(u)\left\{ g'(x)\Delta x + \epsilon_{0}\Delta x\right\} + \epsilon_{1}\left\{ g'(x)\Delta x + \epsilon_{0}\Delta x\right\}[/tex]

    Expanding the parenthesis;

    [tex]\Delta y = f'(u)g'(x)\Delta x + f'(u)\epsilon_{0}\Delta x + \epsilon_{1}g'(x)\Delta x + \epsilon_{1}\epsilon_{0}\Delta x[/tex]

    Dividing throughout by [itex]\Delta x[/itex];

    [tex]\frac{\Delta y}{\Delta x} = f'(u)g'(x) + f'(u)\epsilon_{0} + \epsilon_{1}g'(x) + \epsilon_{1}\epsilon_{0}[/tex]

    Since we want to find the derivative at a point we let [itex]\Delta x\to0[/itex] (recall that [itex]\epsilon\to0[/itex] as [itex]\Delta x\to0[/itex]), thus obtaining;

    [tex]\lim_{\Delta x\to0}\;\left( f'(u)g'(x) + f'(u)\epsilon_{0} + \epsilon_{1}g'(x) + \epsilon_{1}\epsilon_{0}\right)[/tex]

    [tex]=f'(u)g'(x) \stackrel{u=g(x)}{=} f'(g(x))\cdot g(x)[/itex]

    [tex]\therefore{\color{red}\boxed{{\color{black}\frac{dy}{dx}\; f\circ g(x) = \frac{dy}{dx}\; f(g(x)) = f'(g(x))\cdot g'(x) = \frac{dy}{du}\cdot\frac{du}{dx} }}}[/tex]


    'Rough Proof' of chain rule
    from World Web Math, MIT
    This 'proof' is intended for those who have not met differentials and linearization previously, so that they can see that the chain rule has some grounding. This should by no means be considered a correct formal proof.



    Using Leibniz's notation, let [itex] u=2x^2+x[/itex], therefore we have;

    [tex]\frac{d}{dx}\;(2x^2+x)^3 = \left\{ \frac{d}{du} u^3 \right\} \cdot \left\{ \frac{d}{dx} (2x^2+x) \right\}[/tex]

    [tex]=3\cdot u^2 \cdot (4x+1) = 3\cdot(2x^2+x)^2\cdot(4x+1)[/tex]

    Note that the chain rule can be used multiple times to find a derivative.
    Last edited: Nov 10, 2006
  10. Nov 22, 2006 #9


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    A correction in post number four. The formula should read;

    [tex]{\color{red}\boxed{{\color{black}\frac{d}{dx}\left (\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}}}}[/tex]

    As an aside it can also be expressed, perhaps more succinctly, thus;

    [tex]{\color{red}\boxed{{\color{black}\frac{d}{dx}\left (\frac{u}{v}\right)=\frac{v\cdot u' - u\cdot v'}{v^2}}}}[/tex]
  11. Dec 5, 2006 #10


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    Implicit Differentiation
    Up until this point we have always considered functions which can be written explicitly in the form of [itex]y=ax^n + bx^{n-1}...[/itex], these are known an explicit functions. However, some functions or equations can not be written explicitly, instead they must be define in terms of a relationship between two variables, these are known as implicit functions. An example of an implicit function would be that of a unit circle; [itex]x^2 + y^2 =1[/itex]. On occasions it may be possible to write such equations in an explicit form, however, it is often preferable to differentiate implicit functions implicitly as they are.

    When we differentiate implicitly we differentiate the implicit function with respect to the desired variable (x for example); we then treat other variables (y for example) as unknown functions of x and differentiate them accordingly using the chain rule. There is no formula to learn here (except the chain rule obviously), it is simply a technique you must practise.

    Find an expression for the gradient of the unit circle at any given point (x,y)


    The equation for the unit circle in Cartesian coordinates is given by [itex]x^2 + y^2 =1[/itex] . Now, to find the gradient we must differentiate both sides of the equation with respect to x;

    [tex]\frac{d}{dx}\; x^2 + y^2 = \frac{d}{dx}\;1[/tex]

    Now, the derivative of x2 is a simple application of the chain rule; the derivative of 1 is again a simple application of the constant rule, so that leaves us with;

    [tex]2x + \frac{d}{dx}\; y^2 = 0 [/tex]

    Now, since y is some (unknown) function of x; we can apply the chain rule here. (Refresh your knowledge of the chain rule if required) We therefore obtain;

    [tex] 2x + y\cdot\frac{dy}{dx} = 0[/tex]

    [tex]\therefore \;\;\; \frac{dy}{dx} = -\frac{2x}{y}[/tex]
  12. Dec 6, 2006 #11


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    A typo in my previous post. The penultimate line in the example should read as follows;

    [tex] 2x + {\color{red}2}\cdot y\cdot\frac{dy}{dx} = 0[/tex]

    And therefore the final line will read;

    [tex]\therefore \;\;\; m = \frac{dy}{dx} = -\frac{x}{y}[/tex]

    Apologies for any confusion.
  13. Dec 8, 2006 #12
    In the example given for the quotient rule the answer should be, I hope:-


    [tex]\Rightarrow \frac{d}{dx}\frac{x^2+1}{x^2-1} = \frac{2x\cdot(x^2-1) - 2x\cdot(x^2+1)}{(x^2-1)^2} = -\frac{4x}{(x^2-1)^2}[/tex]

    You have no idea how long I puzzled over your answer before I realised that it must be wrong.:smile:
    Last edited: Dec 8, 2006
  14. Dec 8, 2006 #13
    this bit is bugging me for a few reasons.

    firstly as setting h = 0 would make numerator as well as denominator zero, we've no need for the intuitive idea that the expression becomes infinity.

    secondly the bolded limit might be better replaced with expression and the arrow removed, as setting to zero, rather than limiting, makes it no longer a limit and means ideas like +/- infinity are out of place

    an explanation of why the differential is undefined at h = 0 and the need for a limit as a consequence of this might be better
    Last edited: Dec 8, 2006
  15. Dec 12, 2006 #14
    i saw a reply to my point immediately above, but then it was deleted.

  16. Dec 12, 2006 #15


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    Thank you for your incisive comments, you make a valid point. I intend, perhaps sometime after Christmas, to release a 'final' or corrected version of the tutorial in pdf format with additional examples. My intention with posting the 'initial' version on the forums would, as you have done comment on the work and suggest improvements and identify flaws. I thank you again for your useful comments.

    P.S. I removed my previous post as I felt that I didn't give your post sufficient attention.

    You are of course correct Schrodinger, thank you for pointing the error out. I should stop composing post so late in the evening and start using Maple to check my solutions :biggrin:
  17. Dec 12, 2006 #16
    i thought it might be something like that. with all the fuss about 0/0 in the blogsphere i thought it might be nice to show off an area where it's dealt with usefully and rigorously, namely differentiation :)
  18. Dec 12, 2006 #17


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    Differentiation of Trigonometric Functions
    Trigonometric functions has many important applications in science, especially physics. Therefore, it is important that we are able to differentiate equations involving trigonometric functions. Here I intend to illustrate how to differentiate trigonometric functions from first principles and present a series of 'rules' which I would recommend committing to memory if you are going to be dealing with trigonometric functions on a regular basis.

    1. Derivative of Sine
    For no particular reason I will begin with a derivative of a sine function. From out definition of the derivative we have;

    [tex]f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]

    [tex]f(x):=\sin(x) \Rightarrow f'(x) = \lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}[/tex]

    Applying the angle sum identity [itex]{\color{blue}\sin(A+B) = \sin(A)\cos(B)+\cos(A)\sin(B)}[/itex], we obtain;

    [tex]f'(x) = \lim_{h\to0} \frac{\left( \sin(x)\cos(h)+\cos(x)\sin(h) \right)-\sin(x)}{h}[/tex]

    Taking a factor of [itex]\sin(x)[/itex] out we have;

    [tex]f'(x) = \lim_{h\to0} \frac{\sin(x)\left( \cos(h) - 1 \right) + \cos(x)\sin(h)}{h}[/tex]

    Splitting the limit into two separate limits and rewriting the fractions;

    [tex]f'(x) = \lim_{h\to0} \left( \sin(x)\cdot\frac{\cos(h) - 1}{h} \right) + \lim_{h\to0} \left( \cos(x)\cdot\frac{\sin(h)}{h} \right)[/tex]

    [tex]f'(x) = \sin(x)\cdot\lim_{h\to0} \left(\frac{\cos(h) - 1}{h} \right) + \cos(x) \cdot\lim_{h\to0} \left(\frac{\sin(h)}{h} \right)[/tex]

    I state without proof that the following is true. I may prove them at a later time, but for the moment they should be accepted as truths. ([itex]\theta[/itex] in radians)

    [tex]{\color{blue}\boxed{\hspace{1cm}(1)\;\;\lim_{\theta\to0}\frac{\sin\theta}{\theta}= 1 \hspace{1cm}(2)\;\;\lim_{h\to0}\frac{\cos(h)-1}{h}\hspace{1cm}}}[/tex]
    Note that (2) follows directly from (1)

    [tex]\therefore f'(x) = \sin(x)\cdot0 + \cos(x)\cdot1[/tex]

    [tex]{\color{red}\boxed{{\color{black}f'(x) = \cos(x)}}}[/tex]

    As has been alluded to previously, the derivative of a function gives the gradient of that function at any point where that function is differentiable. Therefore, the cosine function represents the rate of change of the sine function. This flash animation authored by David M. Harrison, Dept. of Physics, Univ. of Toronto illustrates this rather neatly. My thanks to EmilK for bringing this resource to my attention.

    Last edited: Dec 12, 2006
  19. Dec 18, 2006 #18


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    2. Derivative of Cosine
    From the definition of the derivative we have;


    [tex]f'(x) = \lim_{h\to0}\frac{\cos(x+h) - \cos(x)}{h}[/tex]

    Invoking the angle sum formula; [itex]\color{blue}\cos(A+B) = \cos(A)\cos(h) - \sin(A)\sin(B)[/itex]; (and splitting the two limits) we obtain;

    [tex]f'(x) = \lim_{h\to0}\left(\frac{\cos(x)\cos(h)-\cos(x)}{h}\right)-\lim_{h\to0}\left(\frac{\sin(x)\sin(h)}{h}\right)[/tex]

    [tex]= \cos(x)\cdot\lim_{h\to0}\left(\frac{\cos(h)-1}{h}\right)-\sin(x)\cdot\lim_{h\to0}\left(\frac{\sin(h)}{h}\right)[/tex]

    As stated above*;

    [tex]{\color{blue}\boxed{\hspace{1cm}(1)\;\;\lim_{\theta\to0}\frac{\sin\theta}{\theta}= 1 \hspace{1cm}(2)\;\;\lim_{h\to0}\frac{\cos(h)-1}{h}=0\hspace{1cm}}}[/tex]


    [tex]f'(x) = \cos(x)\cdot0 - \sin(x)\cdot1[/tex]

    [tex]\therefore {\color{red}\boxed{{\color{black}f'(x) = -\sin(x)}}}[/tex]

    * The blue box has been correct in this post to include the omitted value of the second limit and corrected such that the first limit occurs as [itex]\theta\to0[/itex] NOT [itex]a\to0[/itex]
    Last edited: Dec 19, 2006
  20. Dec 18, 2006 #19


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    Cant wait for you to release the final version of this. Its nice how you took the steps to illustrate how you got the results.

    But when you are deriving sine [and cosine]:

    [tex]{\color{blue}\boxed{\hspace{1cm}(1)\;\;\lim_{\thet a\to0}\frac{\sin\theta}{\theta}= 1 \hspace{1cm}(2)\;\;\lim_{h\to0}\frac{\cos(h)-1}{h}\hspace{1cm}}}[/tex]

    I wish you would indicate why this is so.

    btw, the 4th line of latex text [for cosine] needs reformatting.
    Last edited: Dec 18, 2006
  21. Jan 6, 2007 #20

    Gib Z

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    Umm speaking of Maple, Ive gotten an account on the website, but where do I get the actual program..Its only got supplementary tutorials..I always had the impression it was one program that you open up and you could do calculations in...

    O btw ranger he corrected the blue box in the post just before yours.

    And, Have you seen [tex]\lim_{\x\to0}\sin x= x[/tex] before? I've seen multiple Geometric Proofs for this, which would answer you question. But, Though not actually proving it, It can be seen the tangent at sin x=0 has a gradient one 1, meaning the tangent has an equation y=x. Since, near the tangent, and therefore near zero, the solutions for sin x are the same, leading us to sin x= x for small x, in radians. Dividing both sides by x achieves what you want.
    Last edited: Jan 6, 2007
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