Linearity of differentiation

Let ${\displaystyle a,b\in \mathbb {R} }$. Let ${\displaystyle f,g}$ be functions. Let ${\displaystyle j}$ be a function, where ${\displaystyle j}$ is defined only where ${\displaystyle f}$ and ${\displaystyle g}$ are both defined. (In other words, the domain of ${\displaystyle j}$ is the intersection of the domains of ${\displaystyle f}$ and ${\displaystyle g}$.) Let ${\displaystyle x}$ be in the domain of ${\displaystyle j}$. Let ${\displaystyle j(x)=af(x)+bg(x)}$.

We want to prove that ${\displaystyle j^{\prime }(x)=af^{\prime }(x)+bg^{\prime }(x)}$.

By definition, we can see that

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {\left(af(x+h)+bg(x+h)\right)-\left(af(x)+bg(x)\right)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {af(x+h)+bg(x+h)-af(x)-bg(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {af(x+h)-af(x)+bg(x+h)-bg(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {(af(x+h)-af(x))+(bg(x+h)-bg(x))}{h}}\\&=\lim _{h\rightarrow 0}\left({\frac {af(x+h)-af(x)}{h}}+{\frac {bg(x+h)-bg(x)}{h}}\right)\\&=\lim _{h\rightarrow 0}\left({\frac {a(f(x+h)-f(x))}{h}}+{\frac {b(g(x+h)-g(x))}{h}}\right)\\&=\lim _{h\rightarrow 0}\left(a{\frac {f(x+h)-f(x)}{h}}+b{\frac {g(x+h)-g(x)}{h}}\right)\\\end{aligned}}}$$

In order to use the limits law for the sum of limits, we need to know that ${\textstyle \lim _{h\to 0}a{\frac {f(x+h)-f(x)}{h}}}$ and ${\textstyle \lim _{h\to 0}b{\frac {g(x+h)-g(x)}{h}}}$ both individually exist. For these smaller limits, we need to know that ${\textstyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$ and ${\textstyle \lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}}$ both individually exist to use the coefficient law for limits. By definition, ${\textstyle f^{\prime }(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$ and ${\textstyle g^{\prime }(x)=\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}}$. So, if we know that ${\displaystyle f^{\prime }(x)}$ and ${\displaystyle g^{\prime }(x)}$ both exist, we will know that ${\textstyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$ and ${\textstyle \lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}}$ both individually exist. This allows us to use the coefficient law for limits to write

$${\displaystyle \lim _{h\to 0}a{\frac {f(x+h)-f(x)}{h}}=a\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$$

and

$${\displaystyle \lim _{h\to 0}b{\frac {g(x+h)-f(x)}{h}}=b\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}.}$$

With this, we can go back to apply the limit law for the sum of limits, since we know that ${\textstyle \lim _{h\rightarrow 0}a{\frac {f(x+h)-f(x)}{h}}}$ and ${\textstyle \lim _{h\rightarrow 0}b{\frac {g(x+h)-g(x)}{h}}}$ both individually exist. From here, we can directly go back to the derivative we were working on.

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&\;\;\vdots \\&=\lim _{h\rightarrow 0}\left(a{\frac {f(x+h)-f(x)}{h}}+b{\frac {g(x+h)-g(x)}{h}}\right)\\&=\lim _{h\rightarrow 0}\left(a{\frac {f(x+h)-f(x)}{h}}\right)+\lim _{h\rightarrow 0}\left(b{\frac {g(x+h)-g(x)}{h}}\right)\\&=a\lim _{h\rightarrow 0}\left({\frac {f(x+h)-f(x)}{h}}\right)+b\lim _{h\rightarrow 0}\left({\frac {g(x+h)-g(x)}{h}}\right)\\&=af^{\prime }(x)+bg^{\prime }(x)\end{aligned}}}$$

Finally, we have shown what we claimed in the beginning: ${\displaystyle j^{\prime }(x)=af^{\prime }(x)+bg^{\prime }(x)}$.

Let ${\displaystyle f,g}$ be functions. Let ${\displaystyle j}$ be a function, where ${\displaystyle j}$ is defined only where ${\displaystyle f}$ and ${\displaystyle g}$ are both defined. (In other words, the domain of ${\displaystyle j}$ is the intersection of the domains of ${\displaystyle f}$ and ${\displaystyle g}$.) Let ${\displaystyle x}$ be in the domain of ${\displaystyle j}$. Let ${\displaystyle j(x)=f(x)+g(x)}$.

We want to prove that ${\displaystyle j^{\prime }(x)=f^{\prime }(x)+g^{\prime }(x)}$.

By definition, we can see that

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {\left(f(x+h)+g(x+h)\right)-\left(f(x)+g(x)\right)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {f(x+h)+g(x+h)-f(x)-g(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)+g(x+h)-g(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {(f(x+h)-f(x))+(g(x+h)-g(x))}{h}}\\&=\lim _{h\rightarrow 0}\left({\frac {f(x+h)-f(x)}{h}}+{\frac {g(x+h)-g(x)}{h}}\right)\\\end{aligned}}}$$

In order to use the law for the sum of limits here, we need to show that the individual limits, ${\textstyle \lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$ and ${\textstyle \lim _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}}$ both exist. By definition, ${\textstyle f^{\prime }(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$and ${\textstyle g^{\prime }(x)=\lim _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}}$, so the limits exist whenever the derivatives ${\displaystyle f^{\prime }(x)}$ and ${\displaystyle g^{\prime }(x)}$ exist. So, assuming that the derivatives exist, we can continue the above derivation

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&\;\;\vdots \\&=\lim _{h\rightarrow 0}\left({\frac {f(x+h)-f(x)}{h}}+{\frac {g(x+h)-g(x)}{h}}\right)\\&=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}+\lim _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}\\&=f^{\prime }(x)+g^{\prime }(x)\end{aligned}}}$$

Thus, we have shown what we wanted to show, that: ${\displaystyle j^{\prime }(x)=f^{\prime }(x)+g^{\prime }(x)}$.

Let ${\displaystyle f,g}$ be functions. Let ${\displaystyle j}$ be a function, where ${\displaystyle j}$ is defined only where ${\displaystyle f}$ and ${\displaystyle g}$ are both defined. (In other words, the domain of ${\displaystyle j}$ is the intersection of the domains of ${\displaystyle f}$ and ${\displaystyle g}$.) Let ${\displaystyle x}$ be in the domain of ${\displaystyle j}$. Let ${\displaystyle j(x)=f(x)-g(x)}$.

We want to prove that ${\displaystyle j^{\prime }(x)=f^{\prime }(x)-g^{\prime }(x)}$.

By definition, we can see that:

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {\left(f(x+h)-(g(x+h)\right)-\left(f(x)-g(x)\right)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {f(x+h)-g(x+h)-f(x)+g(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)-g(x+h)+g(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {(f(x+h)-f(x))+(-g(x+h)+g(x))}{h}}\\&=\lim _{h\rightarrow 0}{\frac {(f(x+h)-f(x))-(g(x+h)-g(x))}{h}}\\&=\lim _{h\rightarrow 0}\left({\frac {f(x+h)-f(x)}{h}}-{\frac {g(x+h)-g(x)}{h}}\right)\\\end{aligned}}}$$

In order to use the law for the difference of limits here, we need to show that the individual limits, ${\textstyle \lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$ and ${\textstyle \lim _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}}$ both exist. By definition, ${\textstyle f^{\prime }(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$ and that ${\textstyle g^{\prime }(x)=\lim _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}}$, so these limits exist whenever the derivatives ${\displaystyle f^{\prime }(x)}$ and ${\displaystyle g^{\prime }(x)}$ exist. So, assuming that the derivatives exist, we can continue the above derivation

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&\;\;\vdots \\&=\lim _{h\rightarrow 0}\left({\frac {f(x+h)-f(x)}{h}}-{\frac {g(x+h)-g(x)}{h}}\right)\\&=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}-\lim _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}\\&=f^{\prime }(x)-g^{\prime }(x)\end{aligned}}}$$

Thus, we have shown what we wanted to show, that: ${\displaystyle j^{\prime }(x)=f^{\prime }(x)-g^{\prime }(x)}$.

Let ${\displaystyle f}$ be a function. Let ${\displaystyle a\in \mathbb {R} }$; ${\displaystyle a}$ will be the constant coefficient. Let ${\displaystyle j}$ be a function, where j is defined only where ${\displaystyle f}$ is defined. (In other words, the domain of ${\displaystyle j}$ is equal to the domain of ${\displaystyle f}$.) Let ${\displaystyle x}$ be in the domain of ${\displaystyle j}$. Let ${\displaystyle j(x)=af(x)}$.

We want to prove that ${\displaystyle j^{\prime }(x)=af^{\prime }(x)}$.

By definition, we can see that:

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {af(x+h)-af(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {a\left(f(x+h)-f(x)\right)}{h}}\\&=\lim _{h\rightarrow 0}a{\frac {f(x+h)-f(x)}{h}}\\\end{aligned}}}$$

Now, in order to use a limit law for constant coefficients to show that

$${\displaystyle \lim _{h\rightarrow 0}a{\frac {f(x+h)-f(x)}{h}}=a\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$$

we need to show that ${\textstyle \lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$ exists.

However, ${\textstyle f^{\prime }(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$, by the definition of the derivative. So, if ${\displaystyle f^{\prime }(x)}$ exists, then ${\textstyle \lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$ exists.

Thus, if we assume that ${\displaystyle f^{\prime }(x)}$ exists, we can use the limit law and continue our proof.

$${\displaystyle {\begin{aligned}j^{\prime }(x)&=\lim _{h\rightarrow 0}{\frac {j(x+h)-j(x)}{h}}\\&\;\;\vdots \\&=\lim _{h\rightarrow 0}a{\frac {f(x+h)-f(x)}{h}}\\&=a\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}\\&=af^{\prime }(x)\\\end{aligned}}}$$

Thus, we have proven that when ${\displaystyle j(x)=af(x)}$, we have ${\displaystyle j^{\prime }(x)=af^{\prime }(x)}$.