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8 Oct 2024
Hello everyone! In this blog, we will be reviewing everything about derivatives that you need to know for the AP Calculus test, from the definition of a derivative to more advanced topics like implicit differentiation. This blog is for the curious who want to see the proofs of the differentiation rules we all use. Without further ado, let’s dive into it!
The derivative of a function at a particular point is the slope of the tangent line to the function at that point. We denote a derivative in three ways: $\dfrac{dy}{dx}$, $\dfrac{d}{dx} f(x)$, or $f'(x)$. It is important to note that we are sometimes differentiating with respect to some variable other than $x$ (for example, if we wanted to find the slope of a position-time function, we would differentiate with respect to time). When using $f'(x)$, we are not specifying the differentiation variable, so it is assumed we are deriving with respect to $x$.
To find the derivative of a function $f(x)$ at a point $P(x,f(x))$, we can imagine taking the slope of the line containing $P$ and some other point on the function $Q (x+h, f(x+h))$, where $h$ is the distance in the $x$-coordinate from $P$ to $Q$. We can then find the limit of the slope of $PQ$ as $Q$ moves closer to $P$ (in other words, as $h$ gets smaller). Because slope is $\dfrac{rise}{run}$, this can be notated as
\[ \begin{align*} \dfrac{dy}{dx} = lim_{h\to0} \dfrac{f(x+h) – f(x)}{(x + h) – x} = lim_{h\to0} \dfrac{f(x+h) – f(x)}{h} \end{align*} \]
As an example, let’s take the derivative of $f(x) = x^2$, which is equivalent to
\[ \begin{align*} \lim_{h\to0} \dfrac{(x+h)^2 – x^2}{h} = \lim_{h\to0} \dfrac{x^2 + 2xh + h^2 – x^2}{h} = \lim_{h\to0} \dfrac{2xh + h^2}{h} = \lim_{h\to0} 2x + h = 2x. \end{align*} \]
However, this method can be very slow, especially for more complicated functions. For these, we turn to rules for derivatives, such as a couple of very intuitive rules, the Product Rule, Chain Rule, and Power Rule.
Before we get into more complicated subjects, we first need to lay down a couple of the most basic rules. These are as follows:
$\dfrac{d}{dx} (f+g)(x) = \dfrac{d}{dx} f(x) + \dfrac{d}{dx} g(x) \dfrac{d}{dx} k\cdot f(x) = k \cdot \dfrac{d}{dx} f(x)$
These can both be proven by using the definition of a derivative and limit rules. We will now proceed to more complex (and useful) rules for derivation.
The Product Rule states that $(f(x) \cdot g(x))’ = f'(x)\cdot g(x) + f(x)\cdot g'(x)$. To prove it, let’s use the definition of a derivative:
\[ \begin{align*} (f(x)\cdot g(x))’ = \lim_{h\to 0} \dfrac{f(x+h)g(x+h) – f(x)g(x)}{h} \end{align*} \]
Let us do a clever manipulation by adding $-f(x+h)g(x)+f(x+h)g(x)$ to the numerator. This doesn’t change the value of the limit because we’re adding 0, but it will allow us to evaluate the limit:
\[ \begin{align*} \lim_{h\to 0} \dfrac{f(x+h)g(x+h) – f(x+h)g(x) + f(x+h)g(x) – f(x)g(x)}{h} = \lim_{h\to 0} \dfrac{f(x+h)(g(x+h)-g(x)) + g(x)(f(x+h)-f(x))}{h} =\lim_{h\to 0} \dfrac{f(x+h)(g(x+h)-g(x))}{h} + \lim_{h\to 0} \dfrac{g(x)(f(x+h)-f(x))}{h} &=(\lim_{h\to 0} f(x+h))\left(\lim_{h\to 0} \dfrac{g(x+h)-g(x)}{h}\right) + (\lim_{h\to 0} g(x))\left(\lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\right) \end{align*} \]
Since both $f(x)$ and $g(x)$ are differentiable $lim_{h\to 0} \dfrac{f(x+h) – f(x)}{h} = f'(x)$ and $lim_{h\to 0} \dfrac{g(x+h) – g(x)}{h}= g'(x)$, by definition. Additionally, $lim_{h\to 0} f(x+h)=f(x)$, and $lim_{h\to 0} g(x)=g(x)$. We may thus rewrite the last line as $f(x) g'(x) + f'(x) g(x)$, which is the Product Rule:
$(f(x)\cdot g(x))’ = f'(x) g(x) + f(x) g'(x).$
We will now move on to a very useful rule for differentiating polynomials, the Power Rule.
The Power Rule states that, for any function $f(x) = x^k$, where $k$ is a constant, $f'(x) = kx^{k-1}$. It is extremely important that $k$ is a constant, as many people mistakenly try to apply the Power Rule to functions like $x^{4x}$, where we can’t use the Power Rule because $4x$ is not a constant. We can prove the Power Rule using the Product Rule:
First, we will prove that the derivative of $x$ is $1$. Of course, using our intuitive definition of a derivative as the slope of a function, this becomes obvious, but we can also prove it using the limit definition:
$\lim_{h \to 0} \dfrac{x + h – x}{h} = \lim_{h \to 0} \dfrac{h}{h} = 1$.
Now, we will prove that if the Power Rule holds for some function $f(x) = x^k$, then it will also hold for $g(x) = x^{k+1} = x^k \cdot x^1$. This is a valuable technique called induction. We will use the Product Rule:
$$\dfrac{d}{dx} x^k \cdot x = kx^{k – 1} \cdot x + x^k \cdot 1 = (k+1)x^k,$$ which is exactly what the Product Rule says. For example:
$\dfrac{d}{dx} x^3 = \dfrac{d}{dx} x^2 \cdot x = 2x \cdot x + x^2 = 3x^2$
$\dfrac{d}{dx} x^4 = \dfrac{d}{dx} x^3 \cdot x = 3x^2 \cdot x + x^3 = 4x^3$
and so on.
It is important to recognize that this proof only works when $k$ is a positive integer. The Power Rule does work for all $k$, but proving that requires some more advanced techniques that we will get to in a bit. For now, let’s talk about the Chain Rule.
The Chain Rule states that, for some composite function $f(g(x))$,
$f(g(x))’ = f'(g(x)) \cdot g'(x).$
To find $f'(g(x))$, find $f'(x)$ and then replace $x$ with $g(x)$. The proof is very long and complicated, so we will not put it here. One of the hardest things with the chain rule is identifying which function is the outside function and which is the inside function. Another pitfall is using the Chain Rule when you should use the product rule. For example, $\cos(x) \cdot x^2$ is not a composite function because the two functions are multiplied, so the Product Rule is the correct way to take the derivative. The Chain Rule would be correct for $\cos(x^2)$. We will go over how to take derivatives of trigonometric functions later.
We can now use the Chain Rule to prove the Reciprocal Rule, which states that
\[ \begin{align*} \dfrac{d}{dx} \left(\dfrac{1}{f(x)}\right) = -\dfrac{f'(x)}{f(x)^2}. \end{align*} \]
By the Laws of Exponents, $\dfrac1{f(x)} = f(x)^{-1}$. Applying Chain Rule and Power Rule together,
$\dfrac{d}{dx} \left(\dfrac{1}{f(x)}\right) = \dfrac{d}{dx} f(x)^{-1} = (-1 \cdot f(x)^{-2}) \cdot (f'(x)) = -\dfrac{f'(x)}{f(x)^2}.$
We may now use the Reciprocal Rule to prove the Quotient Rule.
The Quotient Rule states that
$$\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) – f(x)g'(x)}{g(x)^2}.$$
A useful way to remember the Quotient Rule is “low d high minus high d low over low squared,” where “high” is the numerator, “low” is the denominator, and “d” means to take the derivative. We can easily prove this with the Product Rule and Reciprocal Rule:
\[ \begin{align*} &\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{d}{dx} \left(f(x) \cdot \dfrac{1}{g(x)}\right) = f'(x) \cdot \dfrac{1}{g(x)}+ f(x) \left( -\dfrac{g'(x)}{g(x)^2}\right) &=\dfrac{f'(x)g(x)}{g(x)^2} -\dfrac{f(x)g'(x)}{g(x)^2} = \dfrac{f'(x)g(x) – f(x)g'(x)}{g(x)^2}. \end{align*} \]
Before we delve into more advanced topics, let’s quickly cover how to take the derivative of some functions we haven’t covered yet.
The derivative of $\sin(x)$ is $\cos(x)$, and the derivative of $\cos(x)$ is $-\sin(x)$. The derivatives of the other trigonometric functions $\tan(x), \cot(x), \csc(x)$ and $\sec (x)$ can be found by writing them in terms of $\sin (x)$ and $\cos(x)$ and using the Quotient Rule.
We will prove that the derivative of $\sin x$ is $\cos x$ by using $\lim_{x\to 0} \frac{\sin x}{x} = 1$ and $\lim_{x\to 0} \frac{1 – \cos x}{x} = 0$.
Now, we can use the definition of a derivative on $\sin x$:
\[ \begin{align*} \frac{d}{dx} \sin x &= \lim_{h\to 0} \frac{\sin(x+h) – \sin(x)}{h} &= \lim_{h\to0} \dfrac{\sin x\cos h-\cos x\sin h-\sin x}{h} &= \cos x\left(\lim_{h\to0} \frac{\sin h}{h}\right) + \sin x\left(\lim_{h\to0} \frac{1-\cos h}{h}\right) = \cos x. \end{align*} \]
Similarly, for $\cos x$:
\[ \begin{align*} \frac{d}{dx} \cos x &= \lim_{h\to 0} \frac{\cos(x+h) – \cos(x)}{h} &= \lim_{h\to0} \dfrac{\cos x\cos h-\sin x\sin h – \cos x}{h} &= \cos x\left(\lim_{h\to0} \frac{\cos h – 1}{h}\right) – \sin x\left(\lim_{h\to0} \frac{\sin h}{h}\right) = -\sin x. \end{align*} \]
The derivative of $\ln x$ is $\dfrac{1}{x}$, and the derivative of any other logarithmic function can be found by using the Chain Rule. Let’s prove that $\displaystyle\frac d{dx}\ln x = \frac1x$:
\[ \begin{align*} \dfrac{d}{dx} \ln x &= \lim_{h\to 0} \dfrac{\ln(x + h) – \ln(x)}{h} &= \lim_{h\to 0} \dfrac{\ln(\frac{x + h}{x})}{h} &= \lim_{h\to 0} \dfrac{\ln(1 + \frac{h}{x})}{h} \end{align*} \]
We will not prove it here, but $\displaystyle \lim_{t\to 0} \dfrac{\ln(1 + t)}{t} =1$.
It is natural to let $t = \dfrac{h}{x}$, so $h = tx$, and then substitute:
$$\displaystyle \lim_{t\to 0} \dfrac{\ln(1 + t)}{tx}= \dfrac{1}{x} \lim_{t\to 0} \dfrac{\ln(1 + t)}{t} = \frac1x.$$
Now let’s say we wanted to find the derivative of a logarithm to another base, say, $\log_2 x$. We can simply use the change of base formula and then take the derivative:
$$\frac d{dx}\log_2 x=\dfrac{d}{dx} \left(\dfrac{\ln x}{\ln 2}\right) = \dfrac{1}{\ln 2} \cdot \dfrac{d}{dx} \ln x = \dfrac{1}{x\ln 2}.$$
Let us start by finding the derivative of $e^x$ and then we will find derivatives of exponentials with other bases. Somewhat counterintuitively, we can find the derivative of $e^x$ by taking the derivative of $\ln(e^x)$ with the Chain Rule, which is equivalent to $x$:
\[ \begin{align*} \dfrac{d}{dx} \ln(e^x) &= \frac d{dx} x\frac{1}{e^x}\cdot\dfrac{d}{dx} e^x = 1\\ \dfrac{d}{dx} e^x = e^x.\end{align*} \]
If we want to find the derivative of another exponential, say, $2^x$, we can do a bit of manipulation and then use the Chain Rule:
\[ \begin{align*} \dfrac{d}{dx} 2^x = \dfrac{d}{dx}{(e^{\ln 2})}^{x} = \dfrac{d}{dx}e^{x\ln 2}= e^{x \ln 2} \cdot \ln 2 = 2^x \cdot \ln 2. \end{align*} \]
We will now go over the idea of implicit differentiation. This isn’t a formula you can memorize, but instead a very important concept, which can help you take very tricky derivatives.
Implicit differentiation allows you to find derivatives of terms that aren’t defined explicitly in terms of the variable. For example, let us take the equation $x^2 + y^2 = 1$. We can take the derivative of both sides using the Chain Rule ($f(x) = x^2, g(x) = y$):
$2x + 2yy’ = 0 \implies yy’ = -x \implies y’ = -\dfrac{x}{y}$.
The idea is that we use the Chain Rule to get an equation for $y’$ without having to differentiate some function of $x$. As one example of the usefulness of implicit differentiation, let’s go back and prove the Power Rule with all the knowledge we have now.
We want to take the derivative of some function $y = x^k$, where $k$ is constant. Let us take the logarithm and then derivative of both sides:
\[ \begin{align*} &\ln y = \ln (x^k) = k \ln(x) &\dfrac{1}{y} \cdot y’ = \dfrac{k}{x} &y’ = \dfrac{k}{x} \cdot y = k \cdot x^{k-1} .\end{align*} \]
Unlike the previous proof of the Power Rule we showed earlier, this proof does not rely on $k$ being a positive integer.
We can also use implicit differentiation for some other types of derivatives, such as $y =\arcsin x$ by rewriting it as $x = \sin y$ and then taking the derivative:
\[ \begin{align*} &x = \sin y &1 = \cos y\cdot y’ &y’ = \frac{1}{\cos y} = \frac{1}{\sqrt{1-\sin^2 y}}=\frac{1}{\sqrt{1-x^2}}. \end{align*} \]
We can use the same idea to show that $\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}}$ and that $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^2}$.
This is everything about derivatives that you need to know for your calculus class and for the AP test. We hope you learned something from this blog that you can use in your class. Good luck on your future math adventures!
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