MA101.19 The Chain Rule – One Variable

Consider the function f:\mathbb{R} \to \mathbb{R} given by f(x) = \sin{x}. We are used to wring such things as:

i) f'
ii) f'(x)
iii) \displaystyle \frac{df}{dx}.  For example we would write f'(x) = \cos{x} for example.

Equally well, of course, it would be true to write f'(u) = \cos{u}.

The meaning of (i) and (ii) are mathematically precise. f' means the derived function and f'(x) means the value of f' at x.

The meaning of \displaystyle \frac{df}{dx} can be more devious.

It can simply be taken as synonymous with f'(x). That is \displaystyle \frac{df}{dx} = f'(x).

When such is the intention it would be indisputable that \displaystyle \frac{df}{du} means f'(u).

But there are other more shady uses as we will see.

Now consider substituting x = u^2 in f(x) = \sin{x} to define a function F defined as f(u) = \sin{u^2}.

The chain rule says

\begin{aligned} \displaystyle \frac{dF}{du} &= \frac{df}{dx} \cdot \frac{dx}{du} \\&= \cos{x} \cdot 2u \\&= 2u\cos{u^2} \end{aligned}

where here \frac{dF}{du} means F'(u).

Note that F is not equal to f, but mathematicians frequently write the chain rule as,

\displaystyle \frac{df}{du} = \frac{df}{dx} \cdot \frac{dx}{du}.

Here \frac{df}{du} does not mean f'(u) which is after all \cos{u}.

To see the chain rule in a more precise and unambiguous form think of x=u^2 as defining a function g given by g(u)=u^2, then F = f \circ g and we see the chain rule as saying

(f \circ g)'(u) = f'(g(u)) \cdot g'(u)

Of course the u here is an entirely dummy symbol.

MA101.18 Partial Differentiation

Consider the curve in which the surface z=f(x,y) meets the plane y=c a constant.

In this plane z=f(x,c), and \displaystyle \frac{dz}{dx} would be a formula for the gradient of the tangent to the curve.

If we differentiate f(x,y) with respect to x, treating the y as if it was a constant (some say holding y constant), the the derivative obtained is called the partial derivative of f(x,y) with respect to x and we write \displaystyle \frac{\partial f}{\partial x} or f_x or D_x f. Similarily we have \displaystyle \frac{\partial f}{\partial y}.

With functions of more than two variables one differentiates with respect to one of the variables by holding all the other variables constant.

Examples

\displaystyle \frac{\partial}{\partial x} (x^2 y^2 + \tan{x}) = 2xy^2 + \sec{^2}{x}.

\displaystyle \frac{\partial}{\partial y}( x^2 y^2 + tan x) =2x^2 y

\displaystyle \frac{\partial}{\partial x} (2x+\sin{xy}) = 2 + y \cos{xy}

\displaystyle \frac{\partial}{\partial y} (2x+\sin{xy}) = x \cos{xy}

In the obvious way we can have higher order partial derivatives.

\displaystyle \begin{aligned} f(x,y) &= x^2 y + \sin{x} \\ \frac{\partial ^2 f}{\partial x^2} &= \frac{\partial}{\partial x}\bigg[\frac{\partial}{\partial x}(x^2y + \sin{x})\bigg] \\ &= \frac{\partial}{\partial x}\bigg[2xy+\cos{x}\bigg] \\ &= 2y - \sin{x}\end{aligned}

\displaystyle \begin{aligned} \frac{\partial ^2 f}{\partial y \partial x} &= f_{yx} \\ &= \frac{\partial}{\partial y}\bigg[\frac{\partial}{\partial x}(x^2y + \sin{x})\bigg] \\ &= 2x \end{aligned}

\displaystyle \begin{aligned} \frac{\partial ^2 f}{\partial x \partial y} &= f_{xy} \\ &= \frac{\partial}{\partial x}\bigg[\frac{\partial}{\partial y}(x^2y + \sin{x})\bigg] \\ &= 2x \end{aligned}

\displaystyle \begin{aligned} \frac{\partial ^2 f}{\partial y^2} &= 0 \end{aligned}

Similarly for V = \pi r^2 h we have:

\displaystyle \frac{\partial ^2 V}{\partial r^2} = V_{rr} = 2 \pi h

\displaystyle \frac{\partial ^2 V}{\partial r \partial h} = V_{rh} = 2 \pi r

\displaystyle \frac{\partial ^2 V}{\partial h \partial r} = V_{hr} = 2 \pi r

\displaystyle \frac{\partial ^2 V}{\partial h^2} = V_{hh} = 0

For commonly encountered functions f_{xy} we have that

\displaystyle \frac{\partial ^2 f}{\partial x \partial y} = \frac{\partial ^2 f}{\partial y \partial x}

From here on we may assume that all the mixed derivatives are equal.

Note, the normal rules (sum, product, quotient, function of a function) of differentiation apply to partial differentiation.

 

MA101.14 The hyperbolic functions

We define:

\sinh{x}=\frac{1}{2}(e^x-e^{-x})

\cosh{x}=\frac{1}{2}(e^x+e^{-x})

Note that:

\sinh{-x}=\frac{1}{2}(e^{-x}-e^x)=-\sinh{x} – so an odd function.

\cosh{-x}=\frac{1}{2}(e^{-x}+e^x)=\cosh{x} – so an even function.

Also: \cosh{x}-\sinh{x}=e^{-x}\rightarrow 0 as x \to \infty.

We also define:

\begin{aligned} \tanh{x}&=\frac{\sinh{x}}{\cosh{x}} \\ \coth{x}&=\frac{\cosh{x}}{\sinh{x}}, x\neq 0 \\ \mbox{sech }x &=\frac{1}{\cosh{x}} \\ \mbox{cosech }x &=\frac{1}{\sinh{x}}, x\neq 0 \end{aligned}

Note:

\begin{aligned} \tanh{x} &= \frac{e^x-e^{-x}}{e^x+e^{-x}} \\&= \frac{1-e^{-2x}}{1+e^{-2x}} \rightarrow 1 \;\mbox{from below as} x\to\infty \\ &= \frac{e^{2x}-1}{e^{2x}+1} \rightarrow -1 \;\mbox{from above as} x\to -\infty \end{aligned}

Derivative of sinh and cosh

\frac{d(\sinh{x}}{dx}=\frac{1}{2}(e^x+e^{-x})=\cosh{x}

\frac{d(\cosh{x}}{dx}=\frac{1}{2}(e^x-e^{-x})=\sinh{x}

Exercise
Find derivatives of tanh, coth, sech,cosech.

Inverse functions

sinh is an injection and so we have an inverse function denoted by \sinh^{-1}{x}.

Domain \sinh^{-1}{x} = range of \sinh{x} = \mathbb{R}
Range \sinh^{-1}{x} = domain of \sinh{x} = \mathbb{R}

\sinh^{-1}{x} means the real number whose \sinh{x} is x.

tanh is an injection and so we have an inverse function denoted by \tanh^{-1}{x}.

Domain \tanh^{-1}{x} = range of \tanh{x} = (-1,1)
Range \tanh^{-1}{x} = domain of \tanh{x} = \mathbb{R}

\tanh^{-1}{x} means the real number whose \tanh{x} is x.

Note that \cosh{x} is not an injection, so we need a cut down domain to non-negative x’s (x \geq 0) to make it one.

\cosh^{-1}{x} means the non-negative real number whose \cosh{x} is x.

Derivatives of the inverse hyperbolic functions

\begin{aligned} y &= \cosh^{-1}{x} \\ x &= \cosh{y} \\ \frac{dx}{dy}&=\sinh{y} \\ \frac{dy}{dx} &= \frac{1}{\sinh{y}} \\ &= \frac{1}{\sqrt{x^1-1}} \end{aligned}

\mbox{Note: }\forall x: \cosh{^2}{x}-\sinh{^2}{x}=1.

\begin{aligned} y &= \sinh^{-1}{x} \\ x & = \sinh{y} \\ \frac{dx}{dy} &= \cosh{y} \\ \frac{dy}{dx} &= \frac{1}{\cosh{y}}  \\&= \frac{1}{\sqrt{1+x^2}} \end{aligned}

MA101.12 The exponential functions

We define

\exp{x} = 1 +x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+... which in fact converges \forall x \in \mathbb{R}.

\begin{aligned} \exp{x} \cdot  \exp{y} &= (1 +x+\frac{x^2}{2!}+...)(1 +y+\frac{y^2}{2!}+...) \\&= 1 +x+y+\frac{x^2}{2!}+xy+\frac{y^2}{2!}+...+\frac{x^n}{n!}+\frac{x^{n-1}y}{n-1!}+\frac{x^{n-2}y^2}{(n-2)!2!}+...+\frac{y^2}{2!}+... \\&= 1 +(x+y)+\frac{(x+y)^2}{2!}+...+\frac{(x+y)^n}{n!}+... \\&= \exp{(x+y)} \end{aligned}

We have assumed that we can multiply infinite series as if they were finite algebraic expressions. This is not always true, but it is for convergent power series.

The relationship \exp{x}\cdot\exp{y} = \exp{(x+y)} leads us to write \exp{x}=e^x so that e^x\cdot e^y = e^{x+y}.

Note, \exp{1} = e^1 = e = 1+1+\frac{1}{2!}+\frac{1}{3!}+...

On the assumption that a power series can be differentiated term by term \frac{d e^x}{dx} = 0 + 1 +\frac{x^2}{2!}+...=e^x.

Properties of exp

The graph of y=e^x is as shown, (with series definition this is not easy to see), but there is some evidence from the property e^{x_1}\cdot e^{x_2}=e^{x_1+x_2}

Observe from the graph the properties:

  1. \displaystyle \lim_{x\to\infty}e^x=\infty.
  2. \displaystyle \lim_{x\to-\infty}e^x=0.
  3. \forall x, e^x >0.
  4. e^x is continuous everywhere.
  5. D(e^x)=\mathbb{R}.
  6. e^x is an injection (or one to one).
    ie. x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2).

This means there will be an inverse function. We call it the log function. We have \log{y} equal to the unique x such that y=e^x.

y=e^x and x=\ln{y} mean of course the same thing. The graph of y=\ln{x} will be a reflection of y=e^x in the line y=x.

Note that the domain of ln = range of exp = \mathbb{R}^+.

Properties of ln
  1. If y=\ln{x} then x=e^y, so \frac{dx}{dy} = e^y.
    Therefore, \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{e^y} = \frac{1}{x}.
  2. \ln{x} is an injection.
    Proof:
    Let \ln{x}=\ln{y}.
    Put \ln{x}=p and \ln{y}=p.
    So, e^p=x and e^p=y.
    Whence x=y.
  3. \forall x, x=e^{\ln{x}}.
    Proof:
    Let y=e^{ln{x}}.
    Then \ln{y} = \ln{x}.
    Whence x=y by 2.
    So, x=e^{\ln{x}}.
  4. \ln{xy}=\ln{x}+\ln{y}.
    Proof:
    \begin{aligned} \ln{xy}&=\ln{e^{\ln{x}}\cdot e^{\ln{y}}} \\&= \ln{e^{\ln{x}+\ln{y}}} \\ xy &= e^{ln{x}+\ln{y}} \\ \mbox{Hence}\; \ln{x}+\ln{y} &= \ln{xy} \end{aligned}.
  5. \forall a \in \mathbb{R}^+
    (i) if  m in an integer then a^m = m\ln{a}.
    (ii) if \frac{m}{n} is in \mathbb{Q}, then equally we have \ln{a\frac{m}{n}}=\frac{m}{n}\ln{a}. (a^{\frac{m}{n}} \mbox{ means } \sqrt[n]{a^m}.
Definition

For a \in \mathbb{R}^+ and x \in \mathbb{R} we define a^x = e^{x\ln{a}}.

For example 3^{\sqrt{2}}=e^{\sqrt{2}\ln{3}}.

Note that this definition fits in with the usual one when x is rational.

For fixed a \in \mathbb{R}^+, a^x is a function.

Note: \frac{d(a^x)}{dx} = \frac{d(e^{\ln{a^x}}}{dx} = \frac{d(e^{x\ln{a}})}{dx} = \ln{a(e^{x\ln{a}})} = a^x\ln{a}.