# Theorems on Differentiation

Differentiation is used to find the change in the variables. For instance, the rate of change of distance with respect to time can be defined. Theorems on differentiation namely the sum, difference, product and quotient rules are used in solving problems and arriving at the required solution. Differentiation is one of the fundamental concepts in calculus.

## How to Define Differentiation?

Consider a function f (x). The rate of change of the slope of the curve of the given function is called the derivative of the function. The method of finding the derivative is differentiation. Usually, the dependent variable is represented in terms of the independent variable. It is denoted as dy / dx.

## Important Theorems on Differentiation

The following are the important theorems of differentiation.

Consider two functions f (x) and g (x).

1] Sum or difference rule:

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

Sum rule

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

By the definition of derivatives,

$\frac{d}{dx}f(x)+\frac{d}{dx}g(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}+lim_{g\rightarrow 0}\frac{g(x+h)-g(x)}{h}$

Take the RHS,

$RHS=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)+g(x+h)-g(x)}{h}$ $=lim_{h\rightarrow 0}\frac{f(x+h)+g(x+h)-[f(x)+g(x))]}{h}$

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

= LHS

Difference rule

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

By the definition of derivatives,

$\frac{d}{dx}f(x)-\frac{d}{dx}g(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}-lim_{g\rightarrow 0}\frac{g(x+h)-g(x)}{h}$

Take the RHS,

$RHS=lim_{h\rightarrow 0}\frac{[f(x+h)-f(x)]-[g(x+h)-g(x)]}{h}\\$ $=lim_{h\rightarrow 0}\frac{f(x+h)-g(x+h)-[f(x)-g(x))]}{h}\\$ $= lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}-\frac{g(x+h)-g(x)}{h}\\$

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

= LHS

2] Product rule

$\frac{d}{dx}[f(x)*g(x)]=[\frac{d}{dx}f(x)]*g(x)+[\frac{d}{dx}g(x)]*f(x)$

By the definition of derivatives,

$\frac{d}{dx}[f(x)*g(x)]=lim_{h\rightarrow 0}\frac{f(x+h)*g(x+h)-f(x)*(g(x)}{h}$

By using the limit properties,

$lim_{h\rightarrow 0}f(x+h)*\frac{g(x+h)-g(x)}{h}+lim_{h\rightarrow 0}g(x)*\frac{f(x+h)-f(x)}{h}\\$ $[lim_{h\rightarrow 0}f(x+h)]*[lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{h}]+[lim_{h\rightarrow 0}g(x)]*[lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}]\\$ $[\frac{d}{dx}f(x)]*g(x)+[\frac{d}{dx}g(x)]*f(x)\\$

3] Quotient rule

$\frac{d}{dx}=\frac{[\frac{d}{dx}*f(x)]*g(x)-[\frac{d}{dx}*g(x)]*f(x)}{g(x)^2}\\$

By the definition of derivatives,

$lim_{h\rightarrow 0}\frac{\frac{f(x+h))}{g(x+h))}-\frac{f(x))}{g(x))}}{h}\\$

By taking 1 / h as common,

$lim_{h\rightarrow 0}\frac{1}{h}\frac{f(x+h)*g(x)-f(x)*g(x)+f(x)*g(x)-f(x)*g(x+h)}{g(x+h)*g(x)}\\$ $lim_{h\rightarrow 0}\frac{1}{g(x+h)g(x)}=\frac{f(x+h)*g(x)-f(x)*g(x)+f(x)*g(x)-f(x)*g(x+h)}{h}\\$ $[lim_{h\rightarrow 0}\frac{1}{g(x+h)g(x)}]g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x)}{h}\\$

By using the limit properties,

$\frac{1}{lim_{h\rightarrow 0}g(x+h)*g(x)}*([(lim_{h\rightarrow 0}g(x))*lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}]-[(lim_{h\rightarrow 0}f(x))*lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}])\\$ $=\frac{1}{g(x)*g(x)}(g(x)*f'(x)-f(x)*g'(x)\\$

= $\frac{f’g-fg’}{g^2}\\$

4] Constant Rule

$\frac{d}{dx}(c)=0$

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

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=lim_{h\rightarrow 0}\frac{c-c}{h}=lim_{h\rightarrow 0}0=0$

5] Power rule

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

Assume n to be a positive integer.

The binomial theorem states that,

$(a+b)^n=a^n+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}{b^2}+……nab^{n-1}+b^n$

Where,

$(_{n}^{k})=\frac{n!}{k!(n-k)!}$ $n!=n(n-1)(n-2)…..(2)(1)$

From the definition of derivatives and using the binomial theorem,

$f'(x)=lim_{h\rightarrow 0}\frac{(x+h)^n-x^n}{h}\\$ $lim_{h\rightarrow 0}\frac{x^n+nx^{n-1}h+…..}{h}\\$ $f'(x)=lim_{h\rightarrow 0}\frac{nx^{n-1}+\frac{n(n-1)}{2!}x^{n-2}h^2+……nxh^{n-1}+h^n}{h}\\$ $f'(x)=lim_{h\rightarrow 0}{nx^{n-1}}+\frac{n(n-1)}{2!}x^{n-2}h+…..nxh^{n-2}+h^n-1$ $=nx^{n-1}\\$

6] Chain rule

$\frac{d}{dx}f(u(x))=[f’u(x)]\frac{du}{dx}$

7] Rolle’s theorem states that if f is a function that satisfies:

a. f is continuous on the closed interval [a,b],

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

c. f (a) = f (b)

then there exists a point c in the open interval (a,b) such that f'(c) = 0.

8] The mean value theorem is a generalization of Rolle’s theorem which states that if f is a function that satisfies:

a. f is continuous on the closed interval [a,b], and

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

then there exists a point c in the open interval (a,b) such that f'(c) = [f(b) − f(a)] / [b−a] where the right-hand side is the slope of the line connecting the points (a,f(a)) and (b,f(b)). The Mean Value Theorem can be derived from Rolle’s Theorem by considering the function g(x) = [f(x)] − [(f(b) − f(a)) (x−a) / b−a].

## Solved Examples

Example 1: Find dy / dx of $f\left(y\right)=\left(2x^3-4x^2\right)\left(3x^5+x^2\right)$

Solution:

$\frac{d}{dx}\left(\left(2x^3-4x^2\right)\left(3x^5+x^2\right)\right)\\$ $\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)’=f\:’\cdot g+f\cdot g'$ $f=2x^3-4x^2,\:g=3x^5+x^2\\$ $\frac{d}{dx}\left(2x^3-4x^2\right)\left(3x^5+x^2\right)+\frac{d}{dx}\left(3x^5+x^2\right)\left(2x^3-4x^2\right)\\$ $\mathrm{Apply\:the\:Sum/Difference\:Rule}:\left(f\pm g\right)’=f\:’\pm g’\\$ $\frac{d}{dx}\left(2x^3\right)-\frac{d}{dx}\left(4x^2\right)\\$ $\frac{d}{dx}\left(2x^3\right)\\$ $\mathrm{Apply\:the\:Power\:Rule}:\frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}$ $=2\cdot \:3x^{3-1}\\$ $=6x^2\\$ $\frac{d}{dx}\left(4x^2\right)\\$ $=4\cdot \:2x^{2-1}\\$ $=8x\\$ $=6x^2-8x\\$ $\frac{d}{dx}\left(3x^5+x^2\right)\\$ $=\frac{d}{dx}\left(3x^5\right)+\frac{d}{dx}\left(x^2\right)\\$ $=15x^4+2x\\$ $=\left(6x^2-8x\right)\left(3x^5+x^2\right)+\left(15x^4+2x\right)\left(2x^3-4x^2\right)\\$

On simplification,

$=48x^7-84x^6+10x^4-16x^3$

Example 2: In the mean value theorem, f (b) − f (a) = (b − a) f′(c) if a = 4, b = 9 and f (x) = √x then what is the value of c?

Solution:

F (x) = √x

∴f (a) = √4 = 2, f (b) = √9 = 3 ;

f′(x) = 1 / 2√x

Also, f′ (c) = [f (b) − f (a)] / [b − a] = [3 − 2] / [9 − 4] = 1 / 5

∴ [1/ 2] √c = 1 / 5

C = 25 / 4 = 6.25

Example 3: If the function f (x) = x3 − 6x2 + ax + b satisfies Rolles theorem in the interval [1, 3] and f′ [(2√ 3 + 1) / √3] = 0, then find the value of a.

Solution:

f (x) = x− 6x+ ax + b

f′(x) = 3x2 − 12x + a

f′(c) = 0

f′(2 + [1 / √3]) = 0

3 (2 + [1 / √3])2 −12 (2 + [1 / √3]) + a = 0

3 (4 + 1 / 3 + 4√3) − 12 (2 + [1 / √3]) + a = 0

12 + 1 + 4√3 − 24 − 4√3 + a = 0

a = 11

Example 4: Rolle’s theorem is not applicable to the function f (x) = |x| defined on [1, 1] because

A) f is not continuous on [ 1, 1]

B) f is not differentiable on (1,1)

C) f (−1) ≠ f (1)

D) f (−1) = f (1) ≠ 0

Solution:

$f(x)=\left\{\begin{matrix} -x,\,\text{when –1}\le x<0 & \\ x \text \ {when}\ \text{0}\le x\le \text{1} & \end{matrix}\right.\\\text \ Clearly \ f(-1)=|-1|=1=f(1) \\ \text \ But \ Rf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|}{h} =\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{h}=1 \\Lf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|-h|}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{-h}=-1\\ \text \ Therefore \ Rf'(0)\ne Lf'(0) \\ \text \ Hence \ it \ is \ not \ differentiable \ on \ (-1,\,\,1).\\$

Example 5: If $f(x)=\cos x,0\le x\le \frac{\pi }{2}$, then the real number c of the mean value theorem is

$A) \ \frac{\pi }{6}\\ B) \ \frac{\pi }{4}\\ C) \ {{\sin }^{-1}}\left( \frac{2}{\pi } \right)\\ D) \ {{\cos }^{-1}}\left( \frac{2}{\pi } \right)\\$

Solution:

We know that $f'(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow f'(c)=\frac{0-1}{\pi /2}=-\frac{2}{\pi } …..(i)\\$

But $f'(x)=-\sin x\Rightarrow f'(c)=-\sin c ….(ii)\\$

From (i) and (ii), we get $-\sin c=-\frac{2}{\pi }\Rightarrow c={{\sin }^{-1}}\left( \frac{2}{\pi } \right).$

Example 6: Let f (x) satisfy all the conditions of mean value theorem in [0, 2]. If f (0) = 0 and $|f'(x)|\,\le \frac{1}{2}$ for all x, in [0, 2] then

$A) \ f(x)\le 2\\ B) \ |f(x)|\le 1\\ C) \ f(x)=2x\\ D) \ f(x)=3 \text \ for \ at \ least \ one \ x \ in \ [0, 2]\\$

Solution:

$\frac{f(2)-f(0)}{2-0}=f'(x)\Rightarrow \frac{f(2)-0}{2}=f'(x) \Rightarrow \frac{df(x)}{dx}=\frac{f(2)}{2}\\\Rightarrow f(x)=\frac{f(2)}{2}x+c \text \ therefore \ f(0)=0\\ \Rightarrow c=0; \text \ therefore f(x)=\frac{f(2)}{2}x …..(i) \\ \text \ Given |f'(x)|\le \frac{1}{2}\Rightarrow \left| \frac{f(2)}{2} \right|\le \frac{1}{2} …..(ii) \\|f(x)|=\left| \frac{f(2)}{2}x \right|=\left| \frac{f(2)}{2} \right||x|\le \frac{1}{2}|x|\ [from (ii)]$

In [0, 2], for maximum $(x=2), |f(x)|\le \frac{1}{2}.\,\,\,2\Rightarrow |f(x)|\le 1.$