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# Properties of derivatives

schedule Aug 12, 2023
Last updated
local_offer
Calculus
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Almost all properties of derivatives can be derived from the formal definition of derivativeslink as well as the properties of limits.

\begin{equation}\label{eq:VygyD4YWBzdgmccCkPN} \begin{aligned} f'(x)&= \lim_{\Delta{x}\to0}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}\\ \end{aligned} \end{equation}
Theorem.

# Derivative of a constant

If $k\in\mathbb{R}$ is a constant, then:

$$\frac{d}{dx}k=0$$

Proof. We define a function $f(x)=k$. Let's now use the formal definition of derivatives:

\begin{align*} \frac{d}{dx}k&= \frac{d}{dx}f(x)\\ &=\lim_{\Delta{x}\to0}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}\\ &=\lim_{\Delta{x}\to0}\frac{k-k}{\Delta{x}}\\ &=\lim_{\Delta{x}\to0}0\\ &=0\\ \end{align*}

Here, we used the fact that $f(x+\Delta{x})=k$ for the third equality. This completes the proof.

Theorem.

# Derivative of a constant times a function

If $k\in\mathbb{R}$ is a constant, then:

$$\frac{d}{dx}\Big(k\cdot{f(x)}\Big) = k\cdot\frac{d}{dx}f(x)$$

Proof. We define a new function:

$$g(x)=k\cdot{f(x)}$$

We then begin with the formal definition of derivatives:

\begin{align*} \frac{d}{dx}\Big(k\cdot{f(x)}\Big)&=\frac{d}{dx}g(x)\\ &=\lim_{\Delta{x}\to0}\frac{g(x+\Delta{x})-g(x)}{\Delta{x}}\\ &=\lim_{\Delta{x}\to0}\frac{k\cdot{f(x+\Delta{x})}-k\cdot{f(x)}}{\Delta{x}}\\ &=\lim_{\Delta{x}\to0}k\cdot \Big(\frac{{f(x+\Delta{x})}-{f(x)}}{\Delta{x}}\Big)\\ &=k\cdot\lim_{\Delta{x}\to0}\Big(\frac{{f(x+\Delta{x})}-{f(x)}}{\Delta{x}}\Big)\\ &=k\cdot\frac{d}{dx}f(x)\\ \end{align*}

Here, we have taken out the constant term $k$ from the limit using the scalar product rule of limitslink. This completes the proof.

Theorem.

# Sum rule

If $f(x)$ and $g(x)$ are differentiable functions, then:

$$\frac{d}{dx} \Big(f(x)+g(x)\Big)= \frac{d}{dx}f(x)+ \frac{d}{dx}g(x)$$

Proof. Let define a new function $z(x)$ like follows:

$$z(x)=f(x)+g(x)$$

We use the formal definition of derivatives:

\begin{align*} \frac{d}{dx} \Big(f(x)+g(x)\Big)&= \frac{d}{dx}z(x)\\ &=\lim_{\Delta{x}\to0}\frac{z(x+\Delta{x})-z(x)}{\Delta{x}}\\ &=\lim_{\Delta{x}\to0}\frac{(f(x+\Delta{x})+g(x+\Delta{x}))-(f(x)+g(x))}{\Delta{x}}\\ &=\lim_{\Delta{x}\to0}\frac{(f(x+\Delta{x})-f(x))+(g(x+\Delta{x})-g(x))}{\Delta{x}}\\ &=\lim_{\Delta{x}\to0}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}+ \lim_{\Delta{x}\to0}\frac{g(x+\Delta{x})-g(x)}{\Delta{x}}\\ &=\frac{d}{dx}f(x)+\frac{d}{dx}g(x) \end{align*}

This completes the proof.

Theorem.

# Chain rule

Let $f(x)$ and $g(x)$ be both differentiable functions and let $F(x)$ be the composite function:

$$F(x)=(f\circ{g})(x)=f(g(x))$$

The derivative of $F(x)$ is:

$$F'(x)=f'(g(x))\cdot{g'(x)}$$

If we let $y=f(g(x))$ and $u=g(x)$, then the chain rule in Leibniz's notation is:

$$\frac{dy}{dx}= \frac{dy}{du}\cdot \frac{du}{dx}$$

Proof. Let $y=f(g(x))$. By definition of derivatives, we know that the derivative of $y$ with respect to $x$ is:

$$\frac{dy}{dx}=\lim_{\Delta{x}\to0} \frac{f(g(x+\Delta{x}))-f(g(x))}{\Delta{x}}$$

Let's now perform some algebraic manipulation:

\begin{equation}\label{eq:EjqqDXU29n2jd594FrM} \begin{aligned}[b] \frac{dy}{dx}&=\lim_{\Delta{x}\to0} \Big( \frac{f(g(x+\Delta{x}))-f(g(x))}{\Delta{x}} \cdot \frac{g(x+\Delta{x})-g(x)}{g(x+\Delta{x})-g(x)} \Big)\\ &= \lim_{\Delta{x}\to0} \Big( \frac{f(g(x+\Delta{x}))-f(g(x))}{g(x+\Delta{x})-g(x)} \cdot \frac{g(x+\Delta{x})-g(x)}{\Delta{x}} \Big) \end{aligned} \end{equation}

Here, we are assuming $g(x+\Delta{x})-g(x)$ does not equal zero - otherwise we will be dividing by zero. Most proofs of the chain rule silently make this assumption, but this is in fact a flawed assumption because $g(x+\Delta{x})-g(x)$ may actually equal zero for certain functions. For instance, consider the case when $g(x)$ is a flat horizontal line. Adding some infinitesimally small amount $\Delta{x}$ to the input $x$ clearly does not change the output, and so $g(x+\Delta{x})-g(x)=0$ in this case.

We therefore have to consider two separate cases:

• when $g(x+\Delta{x})-g(x)=0$.

• when $g(x+\Delta{x})-g(x)\ne0$.

For the first case, we have that $g(x+\Delta{x})=g(x)$. Therefore, we have that:

$$f(g(x))=f(g(x+\Delta{x}))$$

This means that an infinitesimally small change of $\Delta{x}$ to $x$ will not affect the output at all. In other words, the slope of $f$ is flat, that is, the rate of change of $f$ is always equal to zero:

$$\begin{equation}\label{eq:X3WMn3xuZMdfCxegruw} (f\circ{g})'(x)=0 \end{equation}$$

Similarly, we have that:

$$g(x+\Delta{x})=g(x)$$

Again, since an infinitesimally small change in $x$ does not affect the output, the slope of $g$ must also equal zero. This means that the derivative of $g$ with respect to $x$ must be equal to zero:

$$\begin{equation}\label{eq:GFnBMGxZLVkFQ1qokNp} g'(x)=0 \end{equation}$$

From \eqref{eq:X3WMn3xuZMdfCxegruw} and \eqref{eq:GFnBMGxZLVkFQ1qokNp}, we can clearly see that the Chain rule trivially holds:

$$(f\circ{g})'(x)=f'(g(x))\cdot{g'(x)}$$

Let's now consider the second case when $g(x+\Delta{x})\ne{g(x)}$. By the product rule of limitslink, we can distribute the limit in \eqref{eq:EjqqDXU29n2jd594FrM} like so:

$$\begin{equation}\label{eq:KFof5YOVEjfm0e8U5LZ} \frac{dy}{dx} = \lim_{\Delta{x}\to0} \Big( \frac{f(g(x+\Delta{x}))-f(g(x))}{g(x+\Delta{x})-g(x)} \Big)\cdot \lim_{\Delta{x}\to0} \Big(\frac{g(x+\Delta{x})-g(x)}{\Delta{x}} \Big) \end{equation}$$

Notice how the right limit is simply the definition of the derivative of $g$ with respect to $x$, that is:

$$\begin{equation}\label{eq:bKoAZNqIo5r7chSldqk} \lim_{\Delta{x}\to0} \Big(\frac{g(x+\Delta{x})-g(x)}{\Delta{x}} \Big)=g'(x) \end{equation}$$

Let's now examine the left limit in \eqref{eq:KFof5YOVEjfm0e8U5LZ}:

$$\begin{equation}\label{eq:aApjqRePT9fJreejxoV} \lim_{\Delta{x}\to0} \Big( \frac{f(g(x+\Delta{x}))-f(g(x))}{g(x+\Delta{x})-g(x)} \Big) \end{equation}$$

Let's define a new variable $k$ to represent the denominator, that is:

$$\begin{equation}\label{eq:FbkqJwantXmoE193iYc} k=g(x+\Delta{x})-g(x) \end{equation}$$

We can see that as $\Delta{x}$ tends to zero, $k$ would also tend to zero since $g(x+\Delta{x})\approx{g(x)}$. Next, let's rearrange \eqref{eq:FbkqJwantXmoE193iYc} like so:

$$\begin{equation}\label{eq:dPXTyGR83FGoV9YIfbl} g(x+\Delta{x})=k+g(x) \end{equation}$$

We can now express the limit \eqref{eq:aApjqRePT9fJreejxoV} using $k$ instead of $\Delta{x}$ like so:

$$\begin{equation}\label{eq:ZxLSyYqUzOjZ8cM9B0U} \lim_{\Delta{x}\to0} \Big( \frac{f(g(x+\Delta{x}))-f(g(x))}{g(x+\Delta{x})-g(x)} \Big)= \lim_{k\to0} \Big( \frac{f(k+g(x))-f(g(x))}{k} \Big) \end{equation}$$

The key here is to notice that \eqref{eq:ZxLSyYqUzOjZ8cM9B0U} is the definition of the derivative of $f(g(x))$. To make this clear, let's define a new variable $u$ such that $u=g(x)$. Therefore, \eqref{eq:ZxLSyYqUzOjZ8cM9B0U} becomes:

$$\begin{equation}\label{eq:RcHty25PuyXEuylAV6j} \lim_{k\to0} \Big( \frac{f(k+u)-f(u)}{k} \Big) = f'(u) = f'(g(x)) \end{equation}$$

Now, substituting the expression for the two limits in \eqref{eq:KFof5YOVEjfm0e8U5LZ} gives us the chain rule:

$$\begin{equation}\label{eq:kSDVuhwt1kNE7mY5ZaQ} \frac{dy}{dx} = f'(g(x))\cdot{g'(x)} \end{equation}$$

Finally, let's also express the chain rule using Leibniz's notation. Recall that $y=f(g(x))$ and suppose we let $u=g(x)$ again. Let's take the derivative of $y=f(g(x))=f(u)$ with respect to $u$ to get:

$$\begin{equation}\label{eq:nw5ccQkWTBhasxKCEZV} \frac{dy}{du}=f'(u)=f'(g(x)) \end{equation}$$

Next, we take the derivative of both sides of $u=g(x)$ with respect to $x$ to get:

$$\begin{equation}\label{eq:cXz2kXT5P4Qdc5KuDR2} \frac{du}{dx}=g'(x) \end{equation}$$

Substituting \eqref{eq:nw5ccQkWTBhasxKCEZV} and \eqref{eq:cXz2kXT5P4Qdc5KuDR2} into \eqref{eq:kSDVuhwt1kNE7mY5ZaQ} gives us the chain rule in Leibniz's notation:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

This completes the proof of the chain rule.

Theorem.

# Derivative of natural log

The derivative of natural log is its reciprocal:

$$\frac{d}{dx}\ln(x)=\frac{1}{x}$$

Solution. We begin with the formal definition of derivatives and use basic properties of logarithm to simplify:

\begin{equation}\label{eq:fTAFYivb5a5WjPszK3z} \begin{aligned}[b] \frac{d}{dx}\ln(x)&= \lim_{\Delta{x}\to0}\frac{\ln(x+\Delta{x})-\ln(x)}{\Delta{x}}\\ &= \lim_{\Delta{x}\to0} \Big[\ln\Big(\frac{x+\Delta{x}}{x}\Big)\cdot\frac{1}{\Delta{x}}\Big]\\ &= \lim_{\Delta{x}\to0} \Big[\ln\Big(1+\frac{\Delta{x}}{x}\Big)\cdot\frac{1}{\Delta{x}}\Big]\\ &= \lim_{\Delta{x}\to0} \Big[\ln\Big[\Big(1+\frac{\Delta{x}}{x}\Big)^{1/\Delta{x}}\Big]\Big] \end{aligned} \end{equation}

Suppose we define a new variable $n$ like so:

$$\begin{equation}\label{eq:OFlT1tIhdOqqD5E0c1F} n=\frac{\Delta{x}}{x} \end{equation}$$

From \eqref{eq:OFlT1tIhdOqqD5E0c1F}, we know that as $\Delta{x}$ approaches zero, $n$ tends to zero as well. Let's rearrange \eqref{eq:OFlT1tIhdOqqD5E0c1F} to the following form:

$$\begin{equation}\label{eq:lMPx5HbaDo54Eh3drQN} \frac{1}{\Delta{x}}=\frac{1}{nx} \end{equation}$$

Now, substituting \eqref{eq:OFlT1tIhdOqqD5E0c1F} and \eqref{eq:lMPx5HbaDo54Eh3drQN} into \eqref{eq:fTAFYivb5a5WjPszK3z} gives:

\begin{align*} \frac{d}{dx}\ln(x)&=\lim_{n\to0} \Big[\ln\Big[\Big(1+n\Big)^{1/(nx)}\Big]\Big]\\ &=\lim_{n\to0} \Big[\frac{1}{x}\cdot\ln\Big[\Big(1+n\Big)^{1/n}\Big]\Big]\\ &=\frac{1}{x}\cdot\lim_{n\to0} \Big[\ln\Big[\Big(1+n\Big)^{1/n}\Big]\Big] \end{align*}

Now, we use the property of limit of composite functionslink to swap the order of the limit and natural log:

\begin{align*} \frac{d}{dx}\ln(x) &=\frac{1}{x}\cdot\ln\Big[\lim_{n\to0} \Big[\Big(1+n\Big)^{1/n}\Big]\Big] \end{align*}

The term inside the natural log is the formal definitionlink of $e$ and thus we get:

\begin{align*} \frac{d}{dx}\ln(x) &=\frac{1}{x}\cdot\ln(e)\\ &=\frac{1}{x}\cdot1\\ &=\frac{1}{x} \end{align*}

This completes the proof.

Theorem.

# Power rule

The derivative of $y=x^n$ where $n\in\mathbb{R}$ with respect to $x$ is:

$$\frac{dy}{dx}=nx^{n-1}$$

Solution. Suppose we have the following equation:

$$\begin{equation}\label{eq:hsZr1Fc8hHHMZf4UT3X} y=x^n \end{equation}$$

Where $n\in\mathbb{R}$. Taking the natural logarithm on both sides gives:

$$\ln(y)=n\cdot\ln(x)$$

Now, we are going to perform implicit differentiation, which involves taking the derivative of both sides with respect to $x$ like so:

\begin{equation}\label{eq:CZGjfAEGwVoamin4gJQ} \begin{aligned}[b] \frac{d}{dx}\Big(\ln(y)\Big)&=\frac{d}{dx}\Big(n\cdot\ln(x)\Big)\\ \frac{1}{y}\cdot{y'}&=n\cdot\frac{1}{x}\\ y'&=\frac{ny}{x} \end{aligned} \end{equation}

Now, we substitute \eqref{eq:hsZr1Fc8hHHMZf4UT3X} into \eqref{eq:CZGjfAEGwVoamin4gJQ} to get:

\begin{align*} y'&=\frac{nx^n}{x}\\ &=nx^{n-1} \end{align*}

This completes the proof.

Theorem.

# Product rule

If $f(x)$ and $g(x)$ are differentiable functions, then their product $P(x)=f(x)\cdot{g(x)}$ is also a differentiable function whose derivative is:

$$P'(x)= f(x)\cdot{g'(x)}+ f'(x)\cdot{g(x)}$$

Proof. We are given that:

$$\begin{equation}\label{eq:JKNwEXxB1iFGkN5wvSO} P(x)=f(x)\cdot{g(x)} \end{equation}$$

Taking the natural log of both sides:

\begin{align*} \ln(P(x))&=\ln[f(x)\cdot{g(x)}]\\ &=\ln[f(x)]+\ln[g(x)] \end{align*}

Now, let's perform implicit differentiation by taking the derivative of both sides with respect to $x$ like so:

\begin{align*} \frac{d}{dx}\ln(P(x))&=\frac{d}{dx}\Big(\ln[f(x)]+\ln[g(x)]\Big)\\ \frac{1}{P(x)}\cdot{P'(x)}&=\frac{d}{dx}\ln[f(x)]+\frac{d}{dx}\ln[g(x)]\\ \frac{1}{P(x)}\cdot{P'(x)}&=\frac{1}{f(x)}\cdot{f'(x)}+\frac{1}{g(x)}\cdot{g'(x)}\\ P'(x)&=P(x)\cdot\Big(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\Big)\\ \end{align*}

Here, for the second step, we relied on the derivative of natural logarithmlink, chain rulelink and sum rule of differentiationlink. Substituting \eqref{eq:JKNwEXxB1iFGkN5wvSO} into the above gives us the product rule:

\begin{align*} P'(x)&=(f(x)\cdot{g(x)})\cdot\Big(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\Big)\\ &=g(x)\cdot{f'(x)}+f(x)\cdot{g'(x)} \end{align*}

This completes the proof.

Theorem.

# Differentiability implies continuity

If $f(x)$ is differentiable at point $x=c$, then $f(x)$ is continuouslink at this point.

Proof. For $f(x)$ to be continuous at point $x=c$, the following conditions of continuitylink must be satisfied:

• $f(c)$ exists.

• $\lim_{x\to{c}}f(x)=f(c)$.

We know that if $f(x)$ is differentiable at point $x=c$, then the following limit exists:

\begin{equation}\label{eq:g2IZUNvydNgUGjRU41s} \begin{aligned} {\color{#ADF0A6}f'(c)}&= {\color{#ADF0A6}\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}} \end{aligned} \end{equation}

The fact that this limit exists implies $f(c)$ exists, which is the first criteria of continuity.

Let's now forget about \eqref{eq:g2IZUNvydNgUGjRU41s} for a moment and focus on the following limit:

$$\lim_{x\to{c}}\Big(f(x)-f(c)\Big)$$

From the product rule of limitslink, we have that:

\begin{align*} \lim_{x\to{c}}\Big(f(x)-f(c)\Big) &=\lim_{x\to{c}}\Big(\frac{f(x)-f(c)}{x-c}(x-c)\Big)\\ &={\color{#ADF0A6}\lim_{x\to{c}} \Big(\frac{f(x)-f(c)}{x-c}\Big)}\cdot\lim_{x\to{c}}(x-c)\\ &={\color{#ADF0A6}f'(c)}\cdot{(0)}\\ &=0 \end{align*}

Here, the green limit exists because we are given that $f(x)$ is differentiable at point $x=c$.

To summarize what we've shown:

$$\lim_{x\to{c}}\Big(f(x)-f(c)\Big)=0$$

By the summation rule of limitslink, we have that:

$$\lim_{x\to{c}}f(x)-\lim_{x\to{c}}f(c)=0$$

Since $f(c)$ is some fixed value, the limit as $x$ approaches $c$ is still equal to $f(c)$ and thus:

$$\lim_{x\to{c}}f(x)-f(c)=0$$

Finally, placing $f(c)$ to the right-hand side gives:

$$\lim_{x\to{c}}f(x)=f(c)$$

This is definition of continuity! Therefore, we have just shown that differentiability implies continuity!

Theorem.

# Quotient rule

If $f(x)$ and $g(x)$ are differentiable functions, then:

$$\frac{d}{dx} \frac{f(x)}{g(x)}= \frac{g(x)\cdot{f'(x)}-f(x)\cdot{g'(x)}} {\big(g(x)\big)^2}$$

Proof. Let's define a new function $h(x)$ like so:

$$\begin{equation}\label{eq:qRXsic0qJKNYM0zRcnb} h(x)= \frac{f(x)}{g(x)} \end{equation}$$

Multiplying both sides by $g(x)$ gives:

$$f(x)= h(x)\cdot{g(x)}$$

Let's now take the derivative of both sides with respect to $x$ like so:

$$\begin{equation}\label{eq:OhQvSpBcsoMLR2Aej6n} f'(x)= h'(x)\cdot{g(x)}+ g'(x)\cdot{h(x)} \end{equation}$$

Here, we have used the product rule of differentiation. The reason we did this is so that we get $h'(x)$, which is what we wish to find the expression of!

Let's now rearrange \eqref{eq:OhQvSpBcsoMLR2Aej6n} to make $h'(x)$ the subject:

$$\begin{equation}\label{eq:dJNwqlRrJHVCGFUnFSy} h'(x)= \frac{f'(x)-g'(x)\cdot{h(x)}} {g(x)} \end{equation}$$

Now, let's substitute the definition of $h(x)$, that is \eqref{eq:qRXsic0qJKNYM0zRcnb}, into \eqref{eq:dJNwqlRrJHVCGFUnFSy} to get:

\begin{align*} h'(x)&= \frac{f'(x)-g'(x)\cdot\frac{f(x)}{g(x)}} {g(x)}\\ &= \frac{g(x)\cdot{f'(x)}-g'(x)\cdot{f(x)}} {\big(g(x)\big)^2}\\ \end{align*}

This completes the proof.

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