Question

find derivative from first principle ax^2+b/x

Answer 1

Here is the required answer. Solve the second function the same way and do it

Answer 2

The derivative of the given function

       $f(x) = (ax^2 + \frac{b}{x} )$  is

      $f'(x) = 2ax - \frac{b}{x^2}$

• Now, finding the derivative of the given function by first principle, we have

       $f(x) = (ax^2 + \frac{b}{x} )$

• The derivative of a function f(x) by first principle is given by

         $f'(x) = \lim_{h \to x\1} \frac{f(h)-f(x)}{h-x}$

         therefore

         $f'(x) = \lim_{h \to x\1} \frac{ah^2+\frac{b}{h} - (ax^2+\frac{b}{x}) }{h-x}$

         $f'(x) = \lim_{h\to x\1} \frac{ah^2-ax^2+\frac{b}{h}-\frac{b}{x} }{h-x}$

         $f'(x) = \lim_{h \to x\1} \frac{a(h^2-x^2)+b(\frac{1}{h}-\frac{1}{x}) }{h-x}$

• Now, using the property

            $a^2 - b^2 = (a-b)(a+b)$, we get

        $f'(x) = \lim_{h \to x\1} \frac{a(h-x)(h+x)-b(\frac{h-x}{hx}) }{h-x}$

• Taking (h-x) common from the numerator, we get

         $f'(x) = \lim_{h \to x\1} \frac{(h-x)(a(h+x)-b(\frac{1}{hx})) }{h-x}$

          $f'(x) = \lim_{h \to x\1} a(x+h)- \frac{b}{hx}$

          $f'(x) = a(x+x)- \frac{b}{x^2}$

         $f'(x) = 2ax- \frac{b}{x^2}$

        Hence we get derivative of f(x)

          $f'(x) = 2ax- \frac{b}{x^2}$