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Differentiate from first principle

1. y= x^2 + 3x

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

1 Answer

5 votes

Answer:

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.

User Jay Dansand
by
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