Definition3.6.1The Definition of a Derivative
The computation of the slope of a tangent line, the instantaneous rate of change of a function,
at x=a can be found from the following limit:
\begin{equation}f'(x)=\lim_{a \to x} \frac {f(x) - f(a)} {x-a}\label{men-4}\tag{3.6.1}\end{equation}With a small adjustment in notation this limit can be rephrased.
The derivative of f(x) with respect to x is the function f'(x) and is defined as:
\begin{equation}f'(x)=\lim_{h \to 0} \frac {f(x+h) - f(x)} {h}\label{men-5}\tag{3.6.2}\end{equation}
Definition3.6.2The Power Rule
If f: \mathbb{R} \rightarrow \mathbb{R} is a function such that f(x) = x^r, and f is differentiable at x, then,
: \begin{equation}f'(x) = rx^{r-1}\label{men-6}\tag{3.6.3}\end{equation}
Definition3.6.3Linear Approximation
Given a twice continuously differentiable function f of one real number variable,
Taylor's theorem for the case n = 1 states that:
\begin{equation*} f(x) = f(a) + f'(a)(x - a) + R_2\ \end{equation*}
where R_2 is the remainder term. The linear approximation is obtained by dropping the remainder:
\begin{equation} f(x) \approx f(a) + f'(a)(x - a)\text{.}\label{men-7}\tag{3.6.4}\end{equation}This is a good approximation for x when it is close enough to a;
since a curve, when closely observed, will begin to resemble a straight line.
Therefore, the expression on the right-hand side is just the equation for the tangent line
to the graph of f at (a,f(a)). For this reason, this process is also called
the tangent line approximation.
1
From first principles find the derivative of \fe{f}{x}=\sqrt{3x+1}
AnswerBoth methods yield \(\fe{\fd{f}}{x}=\frac{3}{2\sqrt{3x+1}}\).
SolutionFirst Method:
\begin{align*}
\fe{\fd{f}}{x}&=\lim_{a\to x}\frac{\fe{f}{x}-\fe{f}{a}}{x-a}\\
&=\lim_{a\to x}\frac{\sqrt{3x+1}-\sqrt{3a+1}}{x-a}\\
&=\lim_{a\to x}\frac{\sqrt{3x+1}-\sqrt{3a+1}}{x-a}\cdot\frac{\sqrt{3x+1}+\sqrt{3a+1}}{\sqrt{3x+1}+\sqrt{3a+1}}\\
&=\lim_{a\to x}\frac{(3x+1)-(3a+1)}{(x-a)\left(\sqrt{3x+1}+\sqrt{3a+1}\right)}\\
&=\lim_{a\to x}\frac{3(x-a)}{(x-a)\left(\sqrt{3x+1}+\sqrt{3a+1}\right)}\\
&=\lim_{a\to x}\frac{3}{\sqrt{3x+1}+\sqrt{3a+1}}\\
&=\frac{3}{\sqrt{3x+1}+\sqrt{3x+1}}\\
&=\frac{3}{2\sqrt{3x+1}}
\end{align*}
Second Method:
\begin{align*}
\fe{\fd{f}}{x}&=\lim_{h\to0}\frac{\fe{f}{x+h}-\fe{f}{x}}{h}\\
&=\lim_{h\to0}\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}\\
&=\lim_{h\to0}\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}\cdot\frac{\sqrt{3(x+h)+1}+\sqrt{3x+1}}{\sqrt{3(x+h)+1}+\sqrt{3x+1}}\\
&=\lim_{h\to0}\frac{\left(3(x+h)+1\right)-(3x+1)}{h\left(\sqrt{3(x+h)+1}+\sqrt{3x+1}\right)}\\
&=\lim_{h\to0}\frac{3h}{h\left(\sqrt{3(x+h)+1}+\sqrt{3x+1}\right)}\\
&=\lim_{h\to0}\frac{3}{\left(\sqrt{3(x+h)+1}+\sqrt{3x+1}\right)}\\
&=\frac{3}{\sqrt{3x+1}+\sqrt{3x+1}}\\
&=\frac{3}{2\sqrt{3x+1}}
\end{align*}
2
Differentiate \frac{x^{2} - 2 \, \sqrt{x}}{x} showing each step and stating which rules are used.
Answer
\(1+\dfrac{1}{\sqrt{x^3}}\)
Solution
\begin{align*}
\fe{f}{x}&=\frac{x^{2} - 2 \, \sqrt{x}}{x}\\
\text{Expand this to:}\\
\fe{f}{x}&=x - \frac{2}{\sqrt{x}}\\
\text{Differentiate each part using the power law: } \de {x^r}x = r x^{r-1}\\
\fe{\fd{f}}{x}&=1+\dfrac{1}{\sqrt{x^3}}
\end{align*}
3
Use Linear Approximation 3.6.3 to calculate \sqrt{99.8}.
Answer
\(9.9000\)
Solution
\begin{align*}
\fe{f}{x}&=\fe{f}{a} + \fe{\fd{f}}{a}(x-a)\\
TBD
\end{align*}
4
Implicit Differentiation
y \cos\left(x\right) = \sin\left(x y\right) + 1
Answer
\(\frac{dy}{dx} = -\frac{y \cos\left(x y\right)}{x \cos\left(x y\right) - \cos\left(x\right)} - \frac{y \sin\left(x\right)}{x \cos\left(x y\right) - \cos\left(x\right)}\)
Solution
\begin{align*}
y \cos\left(x\right)&=1 + \sin\left(x y\right)\\
\text{Differentiate through:}\\
y \frac{d\left(\cos\left(x\right)\right)}{dx} + \frac{dy}{dx}\cos\left(x\right)&=
\frac{d\left(\sin\left(xy\right)\right)}{dx}\\
- y \sin\left(x\right) + \frac{dy}{dx}\cos\left(x\right)&=
\frac{d\left(\sin\left(xy\right)\right)}{dx}
\\
\text{Now set } u = xy\\
\frac{d\left(\sin\left(u\right)\right)}{dx}&= \frac{d\left(\sin\left(u\right)\right)}{du}\frac{du}{dx}\\
&= \cos\left(u\right)\frac{du}{dx}\\
&= \cos\left(u\right)\left(x\frac{dy}{dx} + y\frac{dx}{dx}\right)\\
&= x\cos\left(xy\right)\frac{dy}{dx} + y\cos\left(xy\right)\\
\text{Substituting this into our main equation:}\\
- y \sin\left(x\right) + \frac{dy}{dx}\cos\left(x\right)&=
x\cos\left(xy\right)\frac{dy}{dx} + y\cos\left(xy\right)
\\
\text{Rearranging:}\\
\frac{dy}{dx}&= \frac{y \cos\left(x y\right) + y \sin\left(x\right)}{ \cos\left(x\right) - x \cos\left(x y\right) }
\end{align*}
5
Implicit Differentiation
xy+2x+3x^2 = 4
Answer
\(\fe{\fd{y}}{x} = \frac{y}{x} - \frac{2}{x} - 6\)
Solution
Implicit:
\begin{gather*}
xy+2x+3x^2 = 4\\
\text{Differentiate each part using the product rule for } xy\\
x\frac{dy}{dx} + y\frac{dx}{dx} + 2 + 6x = 0\\
x\frac{dy}{dx} + y + 2 + 6x = 0\\
\text{Rearranging gives:} \\
\frac{dy}{dx} = -\frac{\left(y+2+6x\right)}{x}\\
\text{Rearranging original equation for } y\\
y = \frac{4}{x} -2 -3x\\
\text{Substituting gives } y\\
\frac{dy}{dx} = -\frac{\left(\frac{4}{x} -2 -3x+2+6x\right)}{x}\\
\frac{dy}{dx} = -\frac{4}{x^2}-3
\end{gather*}
Explicit:
\begin{gather*}
xy+2x+3x^2 = 4\\
\text{Solve for } y\\
y = \frac{4}{x} -2 -3x\\
\text{And differentiate } y\\
\frac{dy}{dx} = -\frac{4}{x^2} -3
\end{gather*}
As you can see the results are the same for each method.
The problem that implicit differentiation solves is that it is often difficult or
impossible to rearrange to have \(y\) on the left by its own.
Implicit Differentiation of y \cos\left(x\right) = \sin\left(x y\right) + 1
Implicit Differentiation of xy+2x+3x^2 = 4
Exercise 2.1
Exercise 2.2
Exercise 2.3
Exercise 2.4
Exercise 2.5
Exercise 3.1
Exercise 3.2
Exercise 4
