2.5: Exercises
Exercise \(\PageIndex{1}\)
Show that if a function is differentiable, then it is also continuous.
Exercise \(\PageIndex{2}\)
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 \[\frac{dy}{dx} \, \exp(y) = 1 \;\;\; \Rightarrow \frac{dy}{dx} = \frac{1}{\exp(y)} = \frac{1}{x}.\]
Exercise \(\PageIndex{3}\)
Using the definition of non-natural powers, prove that \[\frac{d}{dx} [x^y] = y x^{y-1}, \quad\mathrm{for}\;\;x \in \mathbb{R}^+, \; y \notin \mathbb{N}.\]
Exercise \(\PageIndex{4}\)
Consider \(f(x) = \tanh(\alpha x)\) .
- Sketch \(f(x)\) versus \(x\) , for two cases: (i) \(\alpha = 1\) and (ii) \(\alpha \gg 1\) .
- Sketch the derivative function \(f'(x)\) for the two cases, based on your sketches in part (A) (i.e., without evaluating the derivative directly).
- Evaluate the derivative function, and verify that the result matches your sketches in part (B).
Exercise \(\PageIndex{5}\)
Prove geometrically that the derivatives of the sine and cosine functions are: \[\frac{d}{dx}\sin(x) = \cos(x), \quad\frac{d}{dx}\cos(x) = -\sin(x).\] Hence, derive their Taylor series, Eqs. (2.2.5) and (2.2.6).
Exercise \(\PageIndex{6}\)
For each of the following functions, derive the Taylor series around \(x = 0\) :
- \(f(x) = \ln\left[\alpha \cos(x)\right]\) , to the first 3 non-vanishing terms.
- \(f(x) = \cos\left[\pi\exp(x)\right]\) , to the first 4 non-vanishing terms.
- \(\displaystyle f(x) = \frac{1}{\sqrt{1 \pm x}}\) , to the first 4 non-vanishing terms. Keep track of the signs (i.e., \(\pm\) versus \(\mp\) ).
Exercise \(\PageIndex{7}\)
For each of the following functions, sketch the graph and state the domains over which the function is differentiable:
-
\(f(x) = |\sin(x)|\)
-
\(f(x) = \left[\tan(x)\right]^2\)
-
\(\displaystyle f(x) = \frac{1}{1-x^2}\)
Exercise \(\PageIndex{8}\)
Let \(\vec{v}(x)\) be a vectorial function which takes an input \(x\) (a number), and gives an output value \(\vec{v}\) that is a 2-component vector. The derivative of this vectorial function is defined in terms of the derivatives of each vector component: \[\vec{v}(x) = \begin{bmatrix}v_1(x) \\ v_2(x)\end{bmatrix} \;\; \Rightarrow \;\; \frac{d\vec{v}}{dx} = \begin{bmatrix}dv_1/dx \\ dv_2/dx\end{bmatrix}.\] Now suppose \(\vec{v}(x)\) obeys the vectorial differential equation \[\frac{d\vec{v}}{dx} = \mathbf{A} \vec{v},\] where \[\mathbf{A} = \begin{bmatrix}A_{11} & A_{12} \\ A_{21} & A_{22}\end{bmatrix}\] is a matrix that has two distinct real eigenvectors with real eigenvalues.
-
How many independent numbers do we need to specify for the general solution?
-
Let
\(\vec{u}\)
be one of the eigenvectors of
\(\mathbf{A}\)
, with eigenvalue
\(\lambda\)
: \[\mathbf{A} \vec{u} = \lambda \vec{u}.\] Show that
\(\vec{v}(x) = \vec{u}\, e^{\lambda x}\)
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, \(\vec{v}\) is a two-component function, so it requires \(2n\) indpendent variables. The differential equation \[\frac{d\vec{v}}{dx} = \mathbf{A} \vec{v}\] has order \(n = 1\) , so a total of 2 independent variables is required for the general solution.
Let \(u\) be an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\) , and suppose that \(\vec{v}(x) = \vec{u}\,e^{\lambda x}\) (note that \(\vec{u}\) itself does not depend on \(x\) ). Then \[\begin{align} \frac{d\vec{v}}{dx} &= \vec{u} \frac{d}{dx}\left(e^{\lambda x}\right) \\ &= \lambda \, \vec{u}\, e^{\lambda x} \\ &= \left(\mathbf{A} \vec{u}\right) e^{\lambda x} \\ &= \mathbf{A} \left(\vec{u} e^{\lambda x}\right) \\ &= \mathbf{A} \vec{v}(x).\end{align}\] Hence, \(\vec{v}(x)\) satisfies the desired differential equation.
Let \(\vec{u}_1\) and \(\vec{u}_2\) be the eigenvectors of \(\mathbf{A}\) , with eigenvalues \(\lambda_1\) and \(\lambda_2\) . The general solutions will be \[\vec{v}(x) = c_1 \vec{u}_1 e^{\lambda_1 x} + c_2 \vec{u}_2 e^{\lambda_2 x},\] where \(c_1\) and \(c_2\) are independent variables.