**How to Find Derivatives Using First Principle :**

Here we are going to see how to find derivatives using first principle

Let f be defined on an open interval I ⊆ R containing the point x_{0}, and suppose that

exists. Then f is said to be differentiable at x_{0} and the derivative of f at x0, denoted by f'(x_{0}) , is given by

For a function y = f(x) defined in an open interval (a, b) containing the point x_{0}, the left hand and right hand derivatives of f at x = h are respectively denoted by f'(h^{-}) and f'(h^{+})

f'(h^{-}) = lim _{h-> 0}^{-}[f(x + h) - f(x)] / h

f'(h^{+}) = lim _{h-> 0}^{+}[f(x + h) - f(x)] / h

provided the limits exist.

**Question 1 :**

Find the derivatives of the following functions using first principle.

**Solution :**

(i) f(x) = 6

f'(x) = lim _{h-> 0}[f(x + h) - f(x)] / h

f(x + h) = 6

f'(x) = lim _{h-> 0}((6) - 6)/h

= lim _{h-> 0 }(0/h)

= 0

(ii) f(x) = -4x + 7

f'(x) = lim _{h-> 0}[f(x + h) - f(x)] / h

f(x + h) = -4(x + h) + 7

f'(x) = lim _{h-> 0}(-4(x + h) + 7 - (-4x + 7))/h

= lim _{h-> 0}(-4x - 4h + 7 + 4x - 7)/h

= lim _{h-> 0 }(-4h/h)

= -4

(ii) f(x) = -x^{2} + 2

f'(x) = lim _{h-> 0}[f(x + h) - f(x)] / h

f(x + h) = -(x + h)^{2} + 2

= - (x^{2} + h^{2} + 2xh) + 2

= - x^{2} - h^{2} - 2xh + 2

f'(x) = lim _{h-> 0}((- x^{2} - h^{2} - 2xh + 2) - (-x^{2} + 2))/h

= lim _{h-> 0}(- x^{2} - h^{2} - 2xh + 2 + x^{2} - 2)/h

= lim _{h-> 0}(- h^{2} - 2xh)/h

= lim _{h-> 0}(- h - 2x)

= -2x

**Question 2 :**

Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?

(i) f(x) = |x - 1|

**Solution :**

If the function is differentiable, then

f'(1^{-}) = f'(1^{+})

f'(1^{-}) = [f(x) - f(1)] / (x - 1)

= [-(x - 1) - 0]/(x - 1)

= -1

f'(1^{+}) = [f(x) - f(1)] / (x - 1)

= [(x - 1) - 0]/(x - 1)

= 1

Hence the given function is not differentiable at x = 1.

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