**Physical Meaning of the term Derivative**

**Contents**hide

Instantaneous rate of change of function w.r.t independent variable. limit (Δy/Δx) Δx→0

**Geometrical Meaning of the term Derivative**

It gives the slope of the tangent drawn to the curve at x=a.

**Tangent- **It is the limiting case of the secant slope of the secant.

**Slope of Secant AB-** m_{AB} = f(a+h) – f(a) ⁄ (a + h – a)

f ’ (a^{+}) = lim (h→0) f(a+h) – f(a) / h = slope of tangent at A from RHS

f ’ (a^{+}) = Right hand derivative at x=a

m_{AB’} = f(a-h) – f(a) ⁄ (a – h – a)

f ’ (a^{–}) = lim (h→0) f(a+h) – f(a) / -h = slope of tangent at A from LHS

f ’ (a^{–}) = Left-hand derivative at x=a

If f ’(a^{+}) = f ’(a^{–}) = a finite quantity, then f(x) is said to be differentiable at x=a; and in this case curve y = f(x) has a unique tangent (Not Shown) of finite slope at x = a.

f ’(a^{+}) = f ’(a^{–}) = f ’(a) = finite slope of the tangent at x = 0.

**THEOREM-** If a function is differentiable at x=a then it is also continuous at x = a.

**NOTE- **

- Differentiability ⇒ Continuity
- Continuity ⁄⇒ Differentiability
- Discontinuity ⇒ Non-Differentiability
- Non-Differentiability ⁄⇒ Discontinuity

* NOTE- * If f ’(a

^{+}) = p and f ’(a

^{–}) = q where ‘p’ and q” are finite values, then

- If p = q then f(x) is differentiable at x=a and continuous at x = a.
- If p ≠ q then f(x) is not differentiable at x = a, but may or may not be continuous at x = a.
- f(x) is not differentiable but continuous at x=a then the geometrical graph of the function has a sharp corner at x = a. For Example- See graph of f(x) = |x|
- If the graph of the function exhibits a sharp corner at x=a, it means unit tangent can’t be drawn at x = a.

**Differentiability in an Interval**

- A function is said to be differentiable in (a, b) if it is differentiable at each and every point of the interval.
- Differentiable at [a, b] if

It is differentiable in (a, b) f’ (a^{+}) = limit(h→0) {(f(a+h)- f(a)) / h} = a finite quantity f’ (a^{–}) = limit(h→0) {(f(b-h)- f(a)) /- h} = a finite quantity

- All trigonometric, logarithmic, exponential and polynomial functions are differentiable in their domain (Inverse Trigonometric Functions are not included).

**Note-**

- Derivative in an interval should be checked at all those points where f(x) may be discontinuous.
- For |f(x)| differentiability should be checked at those points where f(x) = 0.

*For Example-* Consider a differentiable function f(x) whose graph is given as- Fig.- A

|f(x)| → Fig.- B F(x) = 0 at A, B, C At x= A |f(x)| is derivable; but may or may not be derivable at x = B & C.

**Q-> Try to solve these functions?**

(i) f(x) = |x^{3}| (ii) f(x) = |x(x-1)|(iii) f(x) = (x-1)|x^{2} – 3x + 2|

**Important Points to Remember**

- If f(x) and g(x) both are differentiable function at x = a then f(x) ± g(x), f(x)×g(x) and f(x)/g(x)[g(x)≠0] will also be differentiable at x = a.
- If f(x) differentiable and g(x) is not differentiable at x = a then f(x) ± g(x) will not be differentiable but nothing definite can be said about derivative of function f(x) × g(x) and f(x)/g(x)[where g(x) ≠ 0] at x = a.
- If f(x) and g(x) both are not differentiable at x = a then nothing definite can be said about derivative of function f(x) × g(x) and f(x)/g(x) [where g(x)≠0] at x = a.
- A derivative of a continuous function need not be continuous.

Ex- f(x)= Two cases Case-1 f(x)= x^{2}sin(1/x); x≠0 Case-2 f(x)= 0; x=0

- If f(x) is derivable at x = a and f(a)=0 and g(x) is continuous at x = a then product function f(x)g(x) will also be derivable at x =a.
- If f(x) is differentiable at x = a and p(h) and g(h) approaches to zero as h → 0

limit (h→0) f(a+p(h)) – f(a+ g(h))/ p(h)- g(h) = f’(a)

**Determination of functions which are differentiable and satisfying some given functional rule**

**Four Basic Steps**

- Write the expression for f'(x) whether LHD or RHD

**f'(x) = limit(h→0) {f(x+h) – f(x)}/ h**

- Manipulate f(x+h)-f(x) using functional rule and get f'(x) in its simplifying form.
- Integrate both sides and f(x) with the constant of its integration.
- Using a suitable value of ‘x’, get the value of the constant of integration.

Methods of Differentiation

**Symbols used for derivative**

*dy/dx *or* f ‘(x) *or* D(y) *or *y’ *or y_{1 }*→ *First Order Derivative

*d ^{2}y/dx^{2}*

*or*

*f ”(x)*or

*D*or

^{2}(y)*y”*or y

_{2 }

_{ }

*→*Second Order Derivative

**Derivative of some standard Functions**

- d(x
^{n})/dx =nx^{n-1}n ∈ R; x>0^{ } - d(a
^{x})/dx =a^{x }ln a a>0 - d(e
^{x})/dx =e^{x} - d(ln x)/dx = 1/x

**Trigonometric Functions**

- d(sin
*x*)/dx = cos*x* - d(cos
*x*)/dx = -sin*x* - d(sec
*x*)/dx = sec*x*tan*x* - d(cosec
*x*)/dx = -cot*x*cosec*x* - d(tan
*x*)/dx =*sec*^{2 }*x* - d(cot
*x*)/dx =*–*cosec^{2 }*x*

**Inverse Trigonometric Functions**

- d(sin
^{-1}*x*)/dx = 1/(1-x^{2})^{1/2}-1< x <1^{ } - d(cos
^{-1}*x*)/dx = -1/(1-x^{2})^{1/2}-1< x <1^{ } - d(cot
^{-1}*x*)/dx = -1/(1+x^{2}) x∈ R - d(tan
^{-1}*x*)/dx = 1/(1+x^{2}) x∈ R - d(sec
^{-1}*x*)/dx = 1/|x|(x^{2}-1)^{1/2 }|x|>1 - d(cosec
^{-1}*x*)/dx = -1/|x|(x^{2}-1)^{1/2 }|x|>1

**Rules in Differentiation**

**Sum or Difference Rule**

If h (x) = f(x) ± g (x)

then h’ (x) = f ‘(x) ± g’ (x)

**Constant Multiple Rule**

If h (x) = k f(x) where k is any constant h’ (x) = k f ‘(x)

**Product Rule**

h (x) = f(x) × g (x) then h’ (x) = f ‘(x) g (x) + g’ (x) f(x) *Note- (fgh)’ = (fg)h’ + (fh)g’ + (gh)f ‘* *or* *(fhg)’ = {f(gh)’ + g(fh)’ + h(fg)’}/2*

#### **Quotient Rule**

d{f(x)/g(x)}/dx = {f ‘(x) g (x) + g’ (x) f(x)}/g^{2}(x) *Also- d{1/f (x)}/dx = – f ‘(x)/f ^{2}(x)*

**Chain Rule**

If y = f(t); t = g(u); u = h(x) then, dy/dx = dy/dt × dt/du × du/dx = f ‘(t) × g'(u) × h'(x)