]> Elementary Functions

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 1: Differentiation; Section 3: Derivatives; page 3

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Elementary Functions

Implicit functions

Implicit functions of the form  F(x,y) = 0  ( were introduced in Section 2, page 5. Since one can look at  y  as a dependent variable of  x , this general expression can be considered as a function of  x  only, and apply the rules of differentiation on both sides. As a result the derivative  dy/dx  can be obtained as function of  x  and  y .

The following example (called the "Folium of Descartes") will be used to clarify this point:

x 3 3axy+ y 3 =0 (

where  a  is a none-zero constant. Differentiation on both sides gives:

3 x 2 3a( y+x dy dx )+3 y 2 dy dx =0    

to obtain finally

dy dx = ay x 2 y 2 ax (

One should notice that ( is expressed in terms of  x  and  y . Furthermore, the function itself could be (as this one is) multiple valued. All this complicates the analysis of an implicit function. Although we are not going to enter into details, we'll continue with this example just to find the roots and the stationary points.

The roots of ( can be found by substituting  y = 0  there, obtaining :  (x,y) = (0,0) .

The stationary points can be obtained from (, namely

ay x 2 =0    

which should also be consistent with the function (, itself and also the requirement that the denomirator of ( does not vanish. After some elementary manipulations we find the stationary point  (x,y) =  ( a 2 3 ,a 4 3 )  . The point  (0,0)  could be a stationary point also, but the denominator of ( vanishes and the result is uncertain.

Trigonometric functions

For obtaining the derivative of  sinx  we'll first need to prove that

lim φ0 sinφ φ =1 (

which is done graphically at Fig. Limit of sinx/x . The proof was done for  x>0 , but since   sinφ φ = sin( φ ) φ  , the limit ( holds also from the left.

Now we can proceed with the differentiation of  sinx

d dx sinx= lim Δx0 sin( x+Δx )sinx Δx = lim Δx0 sinxcosΔx+cosxsinΔxsinx Δx = = lim Δx0 cosx sinΔx Δx lim Δx0 sinx 1cosΔx Δx =cosx lim Δx0 sinx 1 cos 2 Δx ( 1+cosΔx )Δx = =cosx sinx 2 lim Δx0 sin 2 Δx Δx =cosx sinx 2 lim Δx0 sinΔx=cosx } (

The derivative of  cosx  can be obtained easily from   cosx=sin( x+ π 2 )   (

The other trigonometric functions can be expressed by  sinx  and  cosx  and therefore their derivatives can be obtained by applying the rules of differentiation. From the definition of the derivative it follows that if a function is periodic, so is its derivative, and therefore the derivatives of all the triginometric functions are periodic. Here are the derivatives of the main functions:

{ d dx sinx=cosx d dx cosx=sinx d dx tanx=1+ tan 2 x= 1 cos 2 x d dx cotx=( 1+ cot 2 x )= 1 sin 2 x } (

One can ask the question if the argument of a trigonometric function was expressed in degrees instead of radians, will it change anything in the relation   d dx sinx=cosx  ? If your answer is NO, then ask yourself if the relation ( will be different in this case? If your answer is NO again, then go back to the calculation of this limit by expressing  φ  in degrees and hopefully you'll get it right. This was just to show the importance of the argument's units when differentiating. For instance, if you want to use the units of degrees for the independent variable  β , then instead of  sinβ  you have to write  sin(πβ/180)  and use the rules of (

The derivatives of the inverse trigonometric functions can be obtained from ( by applying the Rule 2 (for inverse functions) from ( . If you want to refresh your knowledge about the inverse trigonometric functions look at Section 2, page 3 .

In case of  y = asinx  and  x = siny

d dx asinx= dy dx = 1 dx dy = 1 cosy = 1 ± 1 sin 2 y =± 1 1 x 2 (

The ± sign we obtained is due to the ambiguity (double valued) of the two branches of asinx . It means that the sign depends on the branch of asinx used. In a similar way one obtains the derivatives of the rest of the relevant inverse functions. The list follows:

{ d dx asinx=± 1 1 x 2 d dx acosx= 1 1 x 2 d dx atanx= 1 1+ x 2 d dx acotx= 1 1+ x 2 } (

Log and exponent

The derivative of  logax  will be obtained from the definition:

d dx log a x= lim Δx0 log a ( x+Δx ) log a x Δx = = lim Δx0 log a ( x+Δx x ) 1 Δx = lim Δx0 1 x log a ( 1+ Δx x ) x Δx = = 1 x log a lim Δx0 ( 1+ Δx x ) x Δx = 1 x log a e } (

In ( the limit of an expression was put equal to the natural base  e  defined previously (, which needs some explanation. This limit can be rewritten as the limit of a sequence:

lim Δx0 ( 1+ Δx x ) x Δx = lim n ( 1+ 1 n ) n =e (

of the type  1. As shown in Appendix 01, this sequence converges. Inquisitive users could try and calculate a few terms. On page 5 ( we'll see another way to calculate the natural base  e .

If at ( the base  a  is chosen as  e , the derivative becomes simpler, and this is the reason for the name "natural base".

Since the exponential function is the inverse of the logarithmic, one can write  y = ax  and  x = logay  and use the rule for differentiation of an inverse function

d dx a x = dy dx = 1 dx dy = y log a e = a x lna (

where the property of the logarithm   log a elna=1   was used.

Here is the summary of the derivatives of the exponential and logarithmic functions:

{ d dx a x = a x lna d dx expx=expx d dx log a x= log a e x d dx lnx= 1 x } (

The remarkable result that the derivative of  expx  is the function itself, is demonstrated at Fig. Derivative of exponent .

The properties of the logarithmic function could be used for obtaining derivatives of expressions including powers and exponents. In case that we are seeking the unknown derivative of a function  f(x) , but we know the derivative of  ln[f(x)]  then since   d dx ln[ f( x ) ]= 1 f( x ) d dx f( x )   one obtains

d dx f( x )=f( x ) d dx ln[ f( x ) ] (

As example let's take the function  f(x) = xx−1

d dx x x1 = x x1 d dx ln( x x1 )= = x x1 d dx [ ( x1 )lnx ]= x x1 ( lnx+ x1 x ) } (

Hyperbolic functions

The hyperbolic functions were discussed at section2 page4. Since they are expressions consisting of exponential functions, it is staightforward to obtain the rules for their differentiation. The list of the rules follows

{ d dx sinhx=coshx d dx coshx=sinhx d dx tanhx= 1 cosh 2 x =1 tanh 2 x d dx cothx= 1 sinh 2 x =1 coth 2 x } (

The derivatives of the inverse hyperbolic functions are obtained the same way as the inverse trigonometric functions. One should notice, that the only one double valued inverse hyperbolic function is acoshx . Their list follows.

{ d dx asinhx= 1 1+ x 2 d dx acoshx=± 1 x 2 1 d dx atanhx= 1 1 x 2 | x |<1 d dx acothx= 1 1 x 2 | x |>1 } (

Parametric functions

If a function is presented in parametric form (introduced at Section 2, page 5), the increments of the function  y  and the variable  x  can be written as dependent of an increment of the parameter denoted by  t :

{ Δx=x( t+Δt )x( t ) Δy=t( t+Δt )y( t ) } (

According to the definition

dy dx = lim Δx0 Δy Δx = lim Δt0 Δy Δx = lim Δt0 Δy Δt 1 Δx Δt = dy dt dx dt (

As an example we'll take the parametric presentation of the cycloid (

{ x=r( φsinφ ) y=r( 1cosφ ) } (

where  r  is a positive constant and  φ  is a parameter. With the aid of ( one can calculate the first derivative:

dx dφ =r( 1cosφ ) dy dφ =rsinφ dy dx = sinφ 1cosφ } (

The numerator of  dy/dx  vanishes for any  φ = nπ  where  n  is an integer, but only for  φ = (2n+1)π  the denominator does not, which corresponds to stationary points.

For the rest  φ = 2nπ  one should rewrite the derivative in an unambiguous form:

sinφ 1cosφ = sinφ 1cosφ     1+cosφ 1+cosφ = 1+cosφ sinφ (

which makes the expression  ±  , negative from the left and positive from the right, which corresponds to a minimum according to the previous page.

We need to calculate the second derivative for studying the stationary points. This is done by using the same rule as ( but applied on the first derivative instead of  y :

d 2 y d x 2 = d dφ dy dx dx dφ = cosφ( 1cosφ ) sin 2 φ ( 1cosφ ) 2 r( 1cosφ )    = = cosφ1 r ( 1cosφ ) 3 = 1 r ( 1cosφ ) 2 = 1 4r } (

The second derivative is negative for the stationary points  φ = (2n+1)π  which makes them maxima.

The parametric presentation of a double (or multiple) valued function comes in handy in particular, since there is no need for dealing separetly with different single valued branches. The reason is that for different branches corresponding to the same value of the independent variable  x , the parameter has different values, which automatically separates the branches.

The following example consisting of an ellipse will clarify the point. In Cartesian coordinates the canonical form of the ellipse and the two branches (denoted by  y1  and  y2) are

x 2 a 2 + y 2 b 2 =1{ y 1 =+ b a a 2 x 2 y 2 = b a a 2 x 2 } (

We know that  y1  and  y2  have a maximum and a minimum respectively at  x = 0 , that can be verified by differentation of the branches.

Another way to do that is by use of the parametric presentation

x=acosθ y=bsinθ }0θ<2π (

Following (

dy dx = dy dθ dx dθ = bcosθ asinθ = bcosθ asinθ (

which vanishes for   θ= π 2 , 3π 2  .

Following the same procedure as of ( we obtain for the second derivative

d 2 y d x 2 = d dθ ( dy dx ) dx dθ = b a 2 sin 3 θ { <0forθ= π 2 >0forθ= 3π 2 } (

with the maximum and minimum at the expected locations.


  1. What are the domains of the following functions?
  2. In which part of the domain are they single valued?
  3. Calculate their derivatives!

Exercise 1.

y=acosh( | x | )

Exercise 2.

y=ln[ x( 2coshxsinhx ) ]

Exercise 3. The radii of an ellipse are a and b (a>b). Find the inserted rectangles that have:

  1. the maximal area
  2. the maximal circumference
  3. If you did not get stationary points for the minima, explain why!
    Get those minimal points!
Hint: parametric form of an ellipse (

Exercise 4. Assume a>0 in the Folium of Descartes ( The curve has also a parametric presentation:

{ x= 3at 1+ t 3 y= 3a t 2 1+ t 3 }
  1. Prove its consistency with (!
  2. Find the stationary points by means of the parametric presentation!
  3. Obtain the second derivative and find the maximum and minimum!

Exercise 5. A physical point is moving on a plane according to:

{ x=bcos( ωt )exp( ct ) y=bsin( ωt )exp( ct ) }
where t is time and b , c and ω are positive constants.
  1. Describe the path and the motion!
  2. Calculate dy/dx as function of t !

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Next topic: page 4 Applications

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