Question

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

Answer 1

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.