# Find dy/dx from first principles if y=2x^2?

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### 2 Answers

The derivative of a function y = f(x) from the first principles is the value of: lim h-->0 [ (f(x + h) - f(x))/h]

Here y = f(x) = 2x^2

lim h-->0 [ (f(x + h) - f(x))/h]

=> lim h-->0 [ (2*(x + h)^2 - 2*x^2)/h]

=> 2*lim h-->0 [ ((x + h)^2 - x^2)/h]

=> 2*lim h-->0 [ ((x + h)^2 - x^2)/(x + h - x)]

expand the numerator using a^2 - b^2 = (a - b)(a + b)

=> 2*lim h-->0 [ ((x + h - x)(x + h + x))/(x + h - x)]

=> 2*lim h-->0 [ (x + h + x)]

substitute h = 0

=> 2*2x

=> 4x

**The required derivative dy/dx = 4x**

dy/dx = lim [f(x+h) - f(x)]/h, if h approaches to 0.

lim [f(x+h) - f(x)]/h = lim [2(x+h)^2 - 2x^2]/h

We'll expand the binomial:

lim [2(x+h)^2 - 2x^2]/h = lim (2x^2 + 4xh + 2h^2 - 2x^2)/h

We'll eliminate like terms inside brackets:

lim (2x^2 + 4xh + 2h^2 - 2x^2)/h = lim (4xh + 2h^2)/h

We'll factorize by 2h the numerator:

lim (4xh + 2h^2)/h = lim 2h*(2x + h)/h

We'll simplify and we'll get:

lim 2h*(2x + h)/h = lim 2*(2x + h)

We'll substitute h by the value of accumulation point:

lim 2*(2x + h) = 4x + 2*0

lim 2*(2x + h) = 4x

**The value of dy/dx from first principles is: dy/dx = 4x.**