Almost all properties of derivatives can be derived from the formal definition of derivativeslink as well as the properties of limits.

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

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

Proof. We define a new function:

We then begin with the formal definition of derivatives:

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

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

We use the formal definition of derivatives:

This completes the proof.

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

Let's now perform some algebraic manipulation:

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:

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:

Similarly, we have that:

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:

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

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:

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

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

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

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:

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

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:

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

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:

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

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

This completes the proof of the chain rule.

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

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

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:

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

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

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

This completes the proof.

Solution. Suppose we have the following equation:

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

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

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

This completes the proof.

Proof. We are given that:

Taking the natural log of both sides:

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

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:

This completes the proof.

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:

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:

From the product rule of limitslink, we have that:

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

To summarize what we've shown:

By the summation rule of limitslink, we have that:

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

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

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

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

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

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

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:

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

This completes the proof.