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How to Find the Differential and the Integral of A Quadratic Formula.

Updated on December 11, 2013

Differentiation; Finding the Differential

Say we are given the formula;

3x^3+5x^2-2x-10

The process of finding the differential of this formula begins with dropping any terms that are without an X variable as this will only raise or lower the shape uniformly and so is not important for how we will want to use these equations.

With the formula that remains;

Take the exponent, multiply it by the term in front of it's respective X, and decrease the exponent number by 1.

So;

3(3)x^(3-1)+5(2)x^(2-1)-2(1)x^(1-1)

becomes,

9x^2+10x-2

This is the differetial of 3x^3+5x^2-2x-10. Calculus will often call for a second derivative to be taken. We would simply follow the same process with the derivative we found and multiplying each term by it's exponent, while decreasing that exponent and dropping any terms without variables.

9x^2+10x-2 becomes;

18x+10


The inverse Procedure; Integrals

Finding an expressions integral is quite literally the inverse, or complete opposite of finding a differential.

Take the same expression we began with;

3x^3+5x^2-2x-10

Except we will add one x to each piece of the expression and then determine what number out front (coefficient) would have to be their to produce the current coefficient once multiplied by the increased numbers of x. So we will literally reverse the process of differentiation.

3x^3+5x^2-2x-10 becomes

3/4x^4+5/3X^3-x^2-10x+C, where C is constant number (without a variable X)


Step One

I thought this simple explanation might help beginning Calculus students to get started, What you later do with the unknown added constant, and with trigonometric functions become vital in the next working steps of calculus, so look for more hubs on this subject.

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