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Worked Examples · Example 4

Q.Find the derivative of y=2x2+3xy = 2x^2 + 3x using the first principle (i.e. using the definition dydx=lim⁡h→0f(x+h)−f(x)h\dfrac{dy}{dx} = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}).

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Expand f(x+h)f(x+h), subtract f(x)f(x), divide by hh, and let h→0h\to0 to obtain the derivative from first principles.

dydx=lim⁡h→0f(x+h)−f(x)h\dfrac{dy}{dx}=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}, where f(x)f(x) is the given function.

Given: y=f(x)=2x2+3xy=f(x)=2x^2+3x.

  1. Compute f(x+h)f(x+h):

f(x+h)=2(x+h)2+3(x+h)=2(x2+2xh+h2)+3x+3h=2x2+4xh+2h2+3x+3hf(x+h)=2(x+h)^2+3(x+h)=2(x^2+2xh+h^2)+3x+3h=2x^2+4xh+2h^2+3x+3h

  1. Subtract f(x)=2x2+3xf(x)=2x^2+3x:

f(x+h)−f(x)=(2x2+4xh+2h2+3x+3h)−(2x2+3x)=4xh+2h2+3hf(x+h)-f(x)=\left(2x^2+4xh+2h^2+3x+3h\right)-\left(2x^2+3x\right)=4xh+2h^2+3h

  1. Divide by hh (factor hh out of the numerator first): …

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