Q:

# Differentiate from first principle 1. y= x^2 + 3x2. y= x^2 + 3x + 8

Accepted Solution

A:
Answer:Differentiation of both the term is $$2x+3$$Step-by-step explanation:As we have to use first principle of derivatives lets recall the formula.$$f(x)= \lim_{h\to 0}\frac{(x+h)-f(x)}{h}$$Solving our eqaution.$$y=x^2+3x$$We will work with $$f(x+h)$$ then $$-f(x)$$ separately then put in the above formula.1.$$f(x+h)=(x+h)^2+3(x+h)$$$$(x^2+h^2+2hx+3x+3h)$$Now $$-f(x)$$$$-f(x)=-x^2-3x$$Plugging the values of both.$$f(x)= \lim_{h\to 0}\frac{(x+h)-f(x)}{h}$$$$f(x)= \lim_{h\to 0}\frac{(x^2+h^2+2hx+3h+3x)- x^2-3x}{h}=\frac{h^2+2hx+3h}{h}$$Taking $$h$$ as common.$$f(x)= \lim_{h\to 0}\frac{h^2+2hx+3h}{h}=h+2x+3$$Putting $$h=0$$Then $$f(x)=2x+3$$ is the final derivative.This will be same for $$y= x^2 + 3x + 8$$ as we have to put $$8$$ only.2.$$f(x+h)=(x+h)^2+3(x+h)+8$$$$(x^2+h^2+2hx+3x+3h)$$Then $$-f(x)=-x^2-3x-8$$Plugging the values of both.$$f(x)= \lim_{h\to 0}\frac{(x+h)-f(x)}{h}$$$$f(x)= \lim_{h\to 0}\frac{(x^2+h^2+2hx+3h+3x+8)- x^2-3x-8}{h}=\frac{h^2+2hx+3h}{h}$$Taking $$h$$ as common.$$f(x)= \lim_{h\to 0}\frac{h^2+2hx+3h}{h}=h+2x+3$$Putting $$h=0$$Then $$f(x)=2x+3$$ is the final derivative.So both the derivatives are same.