Question

Differentiate from first principle

1. y= x^2 + 3x

2. y= x^2 + 3x + 8

Answer 1

Differentiation of both the term is
`2x+3`

Explanation:

As we have to use first principle of derivatives lets recall the formula.

`f(x)= \lim_(h\to 0)((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)((x+h)-f(x))/(h)`

`f(x)= \lim_(h\to 0)((x^2+h^2+2hx+3h+3x)- x^2-3x)/(h)=(h^2+2hx+3h)/(h)`

Taking
`h` as common.

`f(x)= \lim_(h\to 0)(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)((x+h)-f(x))/(h)`

`f(x)= \lim_(h\to 0)((x^2+h^2+2hx+3h+3x+8)- x^2-3x-8)/(h)=(h^2+2hx+3h)/(h)`

Taking
`h` as common.

`f(x)= \lim_(h\to 0)(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.