Basic Properties of Derivatives
Continuity
Theorem (Continuity of differentiable functions) If a function \(f\) is differentiable at point \(a\) then it is continuous at \(a\).
Assume that \(f\) is differentiable at \(a\). Then the limit \(\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\) exists and is equal to \(f^{\prime}(a)\). Then for each \(\varepsilon > 0\) there exists \(\delta > 0\) such that if \(h\in(-\delta, \delta)\) then \(\frac{f(a+h)-f(a)}{h}\in (f^{\prime}(a)-\varepsilon,f^{\prime}(a)+\varepsilon)\). This implies that \[|f(a+h)-f(a)|\leq h\cdot \max\{|f^{\prime}(a)+\varepsilon|,|f^{\prime}(a)-\varepsilon|\}.\] Clearly as \(h\to 0\) we have \(|f(a+h)-f(a)|\to 0\). This means that \(f\) is continuous at \(a\).
Derivative of sum of functions
In this section we will prove that a sum of two differentiable functions is differentiable, and that a scalar multiple of a differentiable function is differentiable. We will also derive the properties for the derivative of a sum and the derivative of a scalar multiple of a function.
Theorem (Derivative of a scalar multiple)
Assume that \(c\in\mathbb R\) and that \(f \colon\mathbb R\to\mathbb R\) is a differentiable function at point \(a\). Then \(g(x)=cf(x)\) is also differentiable at \(a\) and its derivative satisfies \(g^{\prime}(a)=cf^{\prime}(a)\).
Since \[\lim_{h\to0}\frac{g(a+h)-g(a)}h=\lim_{h\to 0}\frac{cf(a+h)-cf(a)}h=c\lim_{h\to0}\frac{f(a+h)-f(a)}h=c^{\prime}f(a),\] the function \(g\) is differentiable and \(g^{\prime}(a)=cf^{\prime}(a)\).
Theorem (Derivative of a sum)
If \(f\) and \(g\) are two differentiable functions at point \(a\), then so is \(h(x)=f(x)+g(x)\) and the derivative of \(h\) at \(a\) can be evaluated as:
\[h^{\prime}(a)=f^{\prime}(a)+g^{\prime}(a).\]
Using the additive property of the limit we obtain:
\begin{eqnarray*}\lim_{\varepsilon\to0}\frac{h(a+\varepsilon)-h(a)}{\varepsilon} &=&\lim_{\varepsilon\to0}\frac{f(a+\varepsilon)+g(a+\varepsilon)-(f(a)+g(a))}{\varepsilon} \\ &=&\lim_{\varepsilon\to0}\frac{f(a+\varepsilon)-f(a)}{\varepsilon}+\lim_{\varepsilon\to0}\frac{g(a+\varepsilon)-g(a)}{\varepsilon}\\ &=&f^{\prime}(a)+g^{\prime}(a).\end{eqnarray*}
This implies that \(h\) is differentiable and \(h^{\prime}(a)=f^{\prime}(a)+g^{\prime}(a)\).
From the previous two theorems we can conclude that \[(f-g)^{\prime}=\Big(f+(-1)\cdot g\Big)^{\prime}=f^{\prime}+(-1)\cdot g^{\prime}=f^{\prime}-g^{\prime}.\]
Product Rule
In this section we will derive the formula for the derivative of a product of two functions.
Theorem (Product rule) If \(f\) and \(g\) are two differentiable functions at \(a\in\mathbb R\) that so is \(h=f\cdot g\) and the following formula holds:
\[h^{\prime}(a)=f^{\prime}(a)\cdot g(a)+f(a)\cdot g^{\prime}(a).\]
\begin{eqnarray*}\lim_{\varepsilon\to0}\frac{h(a+\varepsilon)-h(a)}{\varepsilon}&=&
\lim_{\varepsilon\to 0}\frac{f(a+\varepsilon)\cdot g(a+\varepsilon)-f(a)\cdot g(a)}{\varepsilon}\\ &=&
\lim_{\varepsilon\to 0}\frac{f(a+\varepsilon)\cdot g(a+\varepsilon)-f(a)\cdot g(a+\varepsilon)+f(a)\cdot g(a+\varepsilon)-f(a)\cdot g(a)}{\varepsilon}\\
& = &\lim_{\varepsilon\to 0}\left(\frac{f(a+\varepsilon)-f(a)}\varepsilon\cdot g(a+\varepsilon)+ f(a)\frac{g(a+\varepsilon)-g(a)}\varepsilon
\right) \\ &=& \lim_{\varepsilon\to 0} g(a+\varepsilon)\cdot\lim_{\varepsilon\to 0}\frac{f(a+\varepsilon)-f(a)}\varepsilon+ f(a)\cdot \lim_{\varepsilon \to 0 }\frac{g(a+\varepsilon)-g(a)}\varepsilon. \end{eqnarray*}
We now use the fact that \(g\) is continuous at \(a\) (this is implied by its differentiability).
Hence \(\lim_{\varepsilon\to 0}g(a+\varepsilon)=g(a)\). Thus: \[h^{\prime}(a)= g(a)f^{\prime}(a)+f(a)g^{\prime}(a). \]
Example Find the derivative of the function \(f(x)=x^2\cdot \cos x\).
We can write \(f(x)=u(x)\cdot v(x)\), where \(u(x)=x^2\) and \(v(x)=\cos x\). Then \(u^{\prime}(x)=2x\), \(v^{\prime}(x)=-\sin x\), and \[f^{\prime}(x)=u^{\prime}(x)\cdot v(x)+u(x)\cdot v^{\prime}(x)= 2x\cos x-x^2\sin x.\]
Theorem (Derivative of a reciprocal) If \(f\) is a function differentiable at \(a\) that satisfies \(f(a)\neq 0\), then the function \(g(x)=\frac1{f(x)}\) is also differentiable at \(a\) and satisfies \[g^{\prime}(a)=-\frac{f^{\prime}(a)}{f(a)^2}.\]
\[\lim_{\varepsilon\to 0 }\frac{\frac1{f(a+\varepsilon)}-\frac1{f(a)}}{\varepsilon}=\lim_{\varepsilon\to 0}\frac{f(a)-f(a+\varepsilon)}{\varepsilon f(a)f(a+\varepsilon)}= \frac1{f(a)}\cdot\lim_{\varepsilon \to 0}\frac1{f(a+\varepsilon)}\cdot \lim_{\varepsilon\to 0}\frac{f(a)-f(a+\varepsilon)}{\varepsilon}=-\frac{f^{\prime}(a)}{f(a)^2}.\]
We have used the fact that \(f\) is continuous at \(a\) to conlude that \(\lim_{\varepsilon\to0}\frac1{f(a+\varepsilon)}=\frac1{f(a)}\).
Using the previous two theorems we now establish the following result:
Theorem (Quotient rule)
If \(f\) and \(g\) are function differentiable at \(a\) such that \(g(a)\neq 0\), then the function \(h(x)=\frac{f(x)}{g(x)}\) is differentiable at \(a\) and satisfies: \[h^{\prime}(a)=\frac{f^{\prime}(a)\cdot g(a)-f(a)\cdot g^{\prime}(a)}{g(a)^2}.\]
Let us denote \(w(x)=\frac1{g(x)}\). Then we have \(h(x)=f(x)\cdot w(x)\) hence \(h^{\prime}(a)=f^{\prime}(a)\cdot w(a)+f(a)\cdot w^{\prime}(a)\). Since \(w^{\prime}(a)=-\frac{g^{\prime}(a)}{g(a)^2}\) we obtain \[h^{\prime}(a)=f^{\prime}(a)\cdot \frac1{g(a)}+f(a)\cdot \left(-\frac{g^{\prime}(a)}{g(a)^2}\right)=\frac{f^{\prime}(a)\cdot g(a)-f(a)\cdot g^{\prime}(a)}{g(a)^2}.\]
Practice problems
Problem 1. Determine the derivative of \(f(x)=x^3+3x^2-4x+18\).
Since \(f\) is a sum of several functions, its derivative is the sum of derivatives of summands. For example, if \(f(x)=a(x)+b(x)+c(x)+d(x)\), then \(f^{\prime}(x)=a^{\prime}(x)+b^{\prime}(x)+c^{\prime}(x)+d^{\prime}(x)\). This is easily proven since
\[f^{\prime}(x)=a^{\prime}(x)+(b(x)+c(x)+d(x))^{\prime}=a^{\prime}(x)+b^{\prime}(x)+(c(x)+d(x))^{\prime}=a^{\prime}(x)+b^{\prime}(x)+c^{\prime}(x)+d^{\prime}(x).\]
We have \(f^{\prime}(x)=(x^3+3x^2-4x+18)^{\prime}=3x^2+6x-4.\)
Problem 2. Find the equation of the tangent line to the graph of \(f(x)=x+3\sqrt x\) at the point \((1,4)\).
The slope of \(f\) at \((1,4)\) is \(f^{\prime}(1)\). We have \(f^{\prime}(x)=1+\frac32\cdot \frac1{\sqrt x}\) and \(f^{\prime}(1)=1+\frac32=\frac52\). The tangent line has the equation \(\frac{y-4}{x-1}=\frac52\). This is equivalent to \(y=\frac52x+\frac32\).
Problem 3. Assume that \(f\), \(g\), and \(h\) are three differentiable functions. Derive the formula for the derivative of the function \(u(x)=f(x)g(x)h(x)\).
Let us first denote \(p(x)=f(x)\) and \(q(x)=g(x)h(x)\). Then we have \(u^{\prime}=(pq)^{\prime}=p^{\prime}q+pq^{\prime}=f^{\prime}\cdot (gh)+f\cdot (gh)^{\prime}=f^{\prime}gh+f\cdot (g^{\prime}h+gh^{\prime})=f^{\prime}gh+fg^{\prime}h+fgh^{\prime}\).
Problem 4. Let \(f(x)=\sin x\cdot \cos x+x^3\). Find \(f^{\prime}(x)\).
Let us denote \(u(x)=\sin x\cdot \cos x\), and \(v(x)=x^3\). From \(f(x)=u(x)+v(x)\) we conclude that \(f^{\prime}(x)=u^{\prime}(x)+v^{\prime}(x)\). Hence we need to calculate \(u^{\prime}(x)\) and \(v^{\prime}(x)\). The second one directly follows from the power rule: \(v^{\prime}(x)=3x^2\). The first one follows from the product rule. Let us denote \(p(x)=\sin x\) and \(q(x)=\cos x\). Then \(u^{\prime}(x)=p^{\prime}(x)q(x)+p(x)q^{\prime}(x)=\cos x\cdot \cos x-\sin x\cdot\sin x=\cos^2x-\sin^2x\). Therefore \(f^{\prime}(x)=\cos^2x-\sin^2x+3x^2\).
Problem 5. Let \(f(x)= \frac{x}{\cos x+x^3}\). Find \(f^{\prime}(x)\).
Let us denote \(u(x)=x\), and \(v(x)=\cos x+x^3\). From \(f(x)=\frac{u(x)}{v(x)}\) the quotient rule implies
\[f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^2(x)}.\] We first find \(u^{\prime}(x)=1\), \(v^{\prime}(x)=-\sin x+3x^2\). Hence
\[f^{\prime}(x)=\frac{\cos x+x^3-x\cdot (-\sin x+3x^2)}{(\cos x+x^3)^2}=\frac{\cos x+x\sin x-2x^3}{(\cos x+x^3)^2}.\]