Theorems on Differentiation

Theorems on differentiation specifically the sum, difference, product, and quotient rules are used in solving problems and finding the required solution. Here, we will discuss such fundamental theorems on differentiation with their detailed Proof. Differentiation is a fundamental tool of calculus and is also used as an alternative to the slope of the curve. In this article, we will learn and discuss some important theorems on differentiation. This will be followed by some solved examples for better understanding and clarity.

How to Define Differentiation?

Differentiation is defined as a derivative of a function with respect to an independent variable. Consider $y=f(x)$ be a function of $x$. Then, the rate of change of " $y$ " with respect to any change in " $x$ " is given as: $\dfrac{dy}{dx}$.

Mathematically,

$f(x)=\underset{h \rightarrow 0}{\lim}\dfrac{f(x+h)-f(x)}{h}$

Differentiation Graph

History of Mathematician

There were many mathematicians associated with differentiation. However, the definition of differentiation was given by Issac Newton.

Isaac Newton

Name: Isaac Newton

Born: 4 January 1643

Died: 31 March 1727

Field: Mathematics and Science

Nationality: England

Theorems and Proofs of Differentiation

1. Sum or Difference Rule

$\dfrac{d}{d x} f(x) \pm g(x)=\dfrac{d}{d x} f(x) \pm \dfrac{d}{d x} g(x)$\

2. Sum Rule

The derivative of the sum of two functions is the sum of the derivatives of the functions.

$\dfrac{d}{d x} f(x) \pm g(x)=\dfrac{d}{d x} f(x) \pm \dfrac{d}{d x} g(x)$

Proof:

By the definition of derivatives,

$\dfrac{d}{d x} f(x)+\dfrac{d}{d x} g(x)=\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-f(x)}{h}+\underset{g \rightarrow 0}{\lim} \dfrac{g(x+h)-g(x)}{h}$

Take the RHS,

$\qquad R H S=\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-f(x)+g(x+h)-g(x)}{h} \\ =\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)+g(x+h)-[f(x)+g(x))]}{h} \\ =\dfrac{d}{d x}[f(x)+g(x)] \\ =\text { LHS }$

Hence Proved.

3. Difference Rule

The derivative of the difference between two functions is the difference between the derivatives of the functions.

$\dfrac{d}{d x}[f(x)-g(x)]=\dfrac{d}{d x} f(x)-\dfrac{d}{d x} g(x)$

Proof:

By the definition of derivatives,

$\dfrac{d}{d x} f(x)-\dfrac{d}{d x} g(x)=\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-f(x)}{h}-\underset{g \rightarrow 0}{\lim} \dfrac{g(x+h)-g(x)}{h}$

Take the RHS,

$R H S=\underset{h \rightarrow 0}{\lim} \dfrac{[f(x+h)-f(x)]-[g(x+h)-g(x)]}{h} \\ =\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-g(x+h)-[f(x)-g(x))]}{h} \\=\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-f(x)}{h}-\dfrac{g(x+h)-g(x)}{h} \\$

$=\dfrac{d}{d x}[f(x)-g(x)]$

$=$ LHS

Hence Proved.

4. Product Rule

The derivative of the product of two functions is given by the Product Rule.

$\dfrac{d}{d x}[f(x) g(x)]=\left[\dfrac{d}{d x} f(x)\right] g(x)+\left[\dfrac{d}{d x} g(x)\right] f(x)$

Proof:

By the definition of derivatives,

$\dfrac{d}{d x}[f(x) g(x)]=\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h) g(x+h)-f(x) (g(x)}{h}$

By using the limit properties

$\underset{h \rightarrow 0}{\lim} f(x+h) \dfrac{g(x+h)-g(x)}{h}+\underset{h \rightarrow 0}{\lim} g(x) \dfrac{f(x+h)-f(x)}{h} \\\Longrightarrow\left[\underset{h \rightarrow 0}{\lim} f(x+h)\right] \left[\underset{h \rightarrow 0}{\lim} \dfrac{g(x+h)-g(x)}{h}\right]+\left[\underset{h \rightarrow 0}{\lim} g(x)\right] \\\left[\underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-f(x)}{h}\right] \\=\left[\dfrac{d}{d x} f(x)\right] g(x)+\left[\dfrac{d}{d x} g(x)\right] f(x)$

Hence Proved.

5. Quotient Rule

The derivative of the quotient of two functions is given by the Quotient Rule.

$\dfrac{d}{d x}=\dfrac{\left[\dfrac{d}{d x} f(x)\right] g(x)-\left[\dfrac{d}{d x} g(x)\right] f(x)}{g(x)^{2}}$

Proof:

By the definition of derivatives

$\underset{h \rightarrow 0}{\lim} \dfrac{\dfrac{f(x+h))}{g(x+h))}-\dfrac{f(x))}{g(x))}}{h}$

$\Rightarrow \underset{h \rightarrow 0}{\lim} \dfrac{1}{h} \dfrac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{g(x+h) g(x)} \\\Rightarrow \underset{h \rightarrow 0}{\lim} \dfrac{1}{g(x+h) g(x)} \\=\dfrac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h} \\\Rightarrow\left[\underset{h \rightarrow 0}{\lim} \dfrac{1}{g(x+h) g(x)}\right] g(x) \dfrac{f(x+h)-f(x)}{h}-f(x) \dfrac{g(x+h)-g(x)}{h}$

By using the limit properties

$\Rightarrow \dfrac{1}{\underset{h \rightarrow 0}{\lim} g(x+h) g(x)} \left(\left[\left(\underset{h \rightarrow 0}{\lim} g(x)\right) \underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-f(x)}{h}\right]-\left[\left(\underset{h \rightarrow 0}{\lim} f(x)\right)\right.\right. \\\left.\left. \underset{h \rightarrow 0}{\lim} \dfrac{g(x+h)-g(x)}{h}\right]\right) \\\Rightarrow \dfrac{1}{g(x) g(x)}\left(g(x) f^{\prime}(x)-f(x) g^{\prime}(x)\right.$

$\Rightarrow \dfrac{f^{\prime} g-f g^{\prime}}{g^{2}}$

Hence Proved.

6. Constant Rule

The derivative of the constant functions is zero.

$\dfrac{d}{d x}(c)=0$

Proof:

Here $f(x)=c$, by the definition of the derivative, we have

$f^{\prime}(x)= \underset{h \rightarrow 0}{\lim} \dfrac{f(x+h)-f(x)}{h} \\ \Longrightarrow \underset{h \rightarrow 0}{\lim} \dfrac{c-c}{h} \\ \Longrightarrow \underset{h \rightarrow 0}{\lim} 0=0$

7. Power Rule

$\dfrac{d}{d x} x^{n}=n x^{n-1}$

Proof:

Assume $n$ to be a positive integer.

The binomial theorem states that

$(a+b)^{n}=a^{n}+n a^{n-1} b+\dfrac{n(n-1)}{2 !} a^{n-2} b^{2}+\cdots \ldots n a b^{n-1}+b^{n}$

where,

$\left(\begin{array}{l}k \\n\end{array}\right)=\dfrac{n !}{k !(n-k) !} \\n !=n(n-1)(n-2) \ldots . .(2)(1)$

From the definition of derivatives and using the binomial theorem

$\Rightarrow f^{\prime}(x)=\underset{h \rightarrow 0}{\lim} \dfrac{(x+h)^{n}-x^{n}}{h} \\\Rightarrow \underset{h \rightarrow 0}{\lim} \dfrac{x^{n}+n x^{n-1} h+\cdots \cdot}{h} \\\Rightarrow f^{\prime}(x)=\underset{h \rightarrow 0}{\lim} \dfrac{n x^{n-1}+\dfrac{n(n-1)}{2 !} x^{n-2} h^{2}+\cdots \ldots n x h^{n-1}+h^{n}}{h} \\\Rightarrow f^{\prime}(x)=\underset{h \rightarrow 0}{\lim} n x^{n-1}+\dfrac{n(n-1)}{2 !} x^{n-2} h+\cdots \ldots n x h^{n-2}+h^{n}-1 \\=n x^{n-1}$

Hence Proved.

8. Chain Rule

$\dfrac{d}{d x} f(u(x))=\left[f^{\prime} u(x)\right] \dfrac{d u}{d x}$

9. Rolle’s Theorem

Rolle's Theorem states that if $\mathrm{f}$ is a function that satisfies:

1. $f$ is continuous on the closed interval $\left[a, b\right]$,

2. $f$ is differentiable on the open interval $(a, b)$, and

3. $f(a)=f(b)$

Then there exists a point $c$ in the open interval $(a, b)$ such that $f^{\prime}(c)=0$.

Solved Examples

1: Find $\dfrac{dy}{dx}$ of

$f(y)=\left(3 x^{3}-2 x^{2}\right)\left(4 x^{5}+x^{2}\right)$

Ans:

$\dfrac{d}{d x}\left(\left(3 x^{3}-2 x^{2}\right)\left(4 x^{5}+x^{2}\right)\right)$

Apply the Product Rule: $(f \cdot g)^{\prime}=f^{\prime} \cdot g+f \cdot g^{\prime}$

$\Rightarrow f=3 x^{3}-2 x^{2}, g=4 x^{5}+x^{2} \\\Longrightarrow \dfrac{d}{d x}\left(3 x^{3}-2 x^{2}\right)\left(4 x^{5}+x^{2}\right)+\dfrac{d}{d x}\left(4 x^{5}+x^{2}\right)\left(3 x^{3}-2 x^{2}\right)$

Apply the Sum/Difference Rule: $(f \pm g)^{\prime}=f^{\prime} \pm g^{\prime}$

$\Longrightarrow \dfrac{d}{d x}\left(3 x^{3}\right)-\dfrac{d}{d x}\left(2 x^{2}\right)$

$\Rightarrow \dfrac{d}{d x}\left(3 x^{3}\right)$

Apply the Power Rule: $\dfrac{d}{d x}\left(x^{a}\right)=a \cdot x^{a-1}$

$\Rightarrow 3 \cdot 3 x^{3-1}$

$\Longrightarrow 9 x^{2}$

$\Rightarrow \dfrac{d}{d x}\left(2 x^{2}\right)$

$\Rightarrow 2 \cdot 2 x^{2-1}$

$\Longrightarrow 4 x$

So,

$\Rightarrow \dfrac{d}{d x}\left(3 x^{3}\right)-\dfrac{d}{d x}\left(2 x^{2}\right) \\\Rightarrow 9 x^{2}-4 x \\\Rightarrow \dfrac{d}{d x}\left(4 x^{5}+x^{2}\right) \\\Rightarrow \dfrac{d}{d x}\left(4 x^{5}\right)+\dfrac{d}{d x}\left(x^{2}\right) \\\Rightarrow 20 x^{4}+2 x \\\Rightarrow\left(9 x^{2}-4 x\right)\left(4 x^{5}+x^{2}\right)+\left(20 x^{4}+2 x\right)\left(3 x^{3}-2 x^{2}\right) \\\Rightarrow 96 x^{7}-56 x^{6}+15 x^{4}-8 x^{3}$

2: In the mean value theorem, $f(b)-f(a)=(b-a) f^{\prime}(c)$ if $a=9, b=16$ and $f(x)=\sqrt{x}$ then what is the value of $c$ ?

Ans:

$F(x)=\sqrt{x}$

$\therefore f(a)=\sqrt{9}=3, f(b)=\sqrt{16}=4 ; f^{\prime}(x)=\dfrac{1}{2} \sqrt{x}$

Also, $f^{\prime}(c)=\dfrac{[f(b)-f(a)]} {[b-a]}=\dfrac{[4-3]}{16-9}=\dfrac{1}{7}$

$\therefore \dfrac{1}{2\sqrt{c}}=\dfrac{1}{7} \quad C=\dfrac{49}{4}=12.25$

Example 3: If the function $f(x)=x^{3}-6 x^{2}+a x+b$ satisfies Rolle's theorem in the interval $[1,3]$ and $f^{\prime}\left(\dfrac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$, then find the value of $a$.

Ans:

$f(x)=x^{3}-6 x^{2}+a x+b \\ \Rightarrow f^{\prime}(x)=3 x^{2}-12 x+a \\ \Rightarrow f^{\prime}(c)=0 \\ \Rightarrow f^{\prime}\left(2+\dfrac{1}{\sqrt{3}}\right)=0 \\ \Rightarrow 3(2+\sqrt{3})^{2}-12\left(2+\dfrac{1}{\sqrt{3}}\right)+a=0 \\ \Rightarrow 3\left(4+\dfrac{1}{3}+4 \sqrt{3}\right)-12\left(2+\dfrac{1}{\sqrt{3}}\right)+a=0 \\ \Rightarrow 12+1+4 \sqrt{3}-24-4 \sqrt{3}+a=0$

Now we get

$\Rightarrow a=11$

Conclusion

In this article, we have discussed the theorems on differentiation and their proofs. The proof, statement, and examples are thoroughly explained above. From the discussion above about the theorem on differentiation, we can conclude that they are very useful theorems to be used in solving calculus questions.

Important Differentiation Formulas to Remember

• $\dfrac{d k}{d x}=0 \quad$ where $k=$ constant

• $\dfrac{d(x)}{d x}=1$

• $\dfrac{d(k x)}{d x}=k \quad$ where $k=$ constant

• $\dfrac{d\left(x^{n}\right)}{d x}=n x^{n-1}$

Important Points to Remember

• Concept of Differentiation, how to define differentiation must be studied in depth.

• Differentiation is a synonym of slope.