Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Physics LibreTexts

2.5: Exercises

( \newcommand{\kernel}{\mathrm{null}\,}\)

Exercise 2.5.1

Show that if a function is differentiable, then it is also continuous.

Exercise 2.5.2

Prove that the derivative of ln(x) is 1/x.

Answer

If y=ln(x), it follows from the definition of the logarithm that exp(y)=x. Taking d/dx on both sides, and using the product rule, gives dydxexp(y)=1dydx=1exp(y)=1x.

Exercise 2.5.3

Using the definition of non-natural powers, prove that ddx[xy]=yxy1,forxR+,yN.

Exercise 2.5.4

Consider f(x)=tanh(αx).

  1. Sketch f(x) versus x, for two cases: (i) α=1 and (ii) α1.
  2. Sketch the derivative function f(x) for the two cases, based on your sketches in part (A) (i.e., without evaluating the derivative directly).
  3. Evaluate the derivative function, and verify that the result matches your sketches in part (B).

Exercise 2.5.5

Prove geometrically that the derivatives of the sine and cosine functions are: ddxsin(x)=cos(x),ddxcos(x)=sin(x). Hence, derive their Taylor series, Eqs. (2.2.5) and (2.2.6).

Exercise 2.5.6

For each of the following functions, derive the Taylor series around x=0:

  1. f(x)=ln[αcos(x)], to the first 3 non-vanishing terms.
  2. f(x)=cos[πexp(x)], to the first 4 non-vanishing terms.
  3. f(x)=11±x, to the first 4 non-vanishing terms. Keep track of the signs (i.e., ± versus ).

Exercise 2.5.7

For each of the following functions, sketch the graph and state the domains over which the function is differentiable:

  1. f(x)=|sin(x)|

  2. f(x)=[tan(x)]2

  3. f(x)=11x2

Exercise 2.5.8

Let v(x) be a vectorial function which takes an input x (a number), and gives an output value v that is a 2-component vector. The derivative of this vectorial function is defined in terms of the derivatives of each vector component: v(x)=[v1(x)v2(x)]dvdx=[dv1/dxdv2/dx]. Now suppose v(x) obeys the vectorial differential equation dvdx=Av, where A=[A11A12A21A22] is a matrix that has two distinct real eigenvectors with real eigenvalues.

  1. How many independent numbers do we need to specify for the general solution?

  2. Let u be one of the eigenvectors of A, with eigenvalue λ: Au=λu. Show that v(x)=ueλx is a specific solution to the vectorial differential equation. Hence, find the general solution.

Answer

For an ordinary differential equation for a scalar (one-component) function of order n, the general solution must contain n independent variables. In this case, v is a two-component function, so it requires 2n indpendent variables. The differential equation dvdx=Av has order n=1, so a total of 2 independent variables is required for the general solution.

Let u be an eigenvector of A with eigenvalue λ, and suppose that v(x)=ueλx (note that u itself does not depend on x). Then dvdx=uddx(eλx)=λueλx=(Au)eλx=A(ueλx)=Av(x). Hence, v(x) satisfies the desired differential equation.

Let u1 and u2 be the eigenvectors of A, with eigenvalues λ1 and λ2. The general solutions will be v(x)=c1u1eλ1x+c2u2eλ2x, where c1 and c2 are independent variables.


This page titled 2.5: Exercises is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?