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DIFFERENTIATION FROM FIRST PRINCIPLES WJEC C1 PAST PAPER QUESTION January 2008

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Given that Find From first Principles Remember that this is the DERIVATIVE. We are asked to DIFFERENTIATE using FIRST PRINCIPLES REMEMBER that the DERIVATIVE means the GRADIENT OF THE TANGENT to a curve. Here we will end up with an algebraic expression because we do not know the specific point to find the gradient.

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The formula that we will use is To obtainwe substituteWherever we see the x in the function This is our function that we wish to differentiate

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Expand brackets and simplify WORKINGS This will be the SAME in EVERY Case!!! EXPAND BRACKETS This can NOT simplify so DON’T EVEN TRY!!!

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Consider the gradient of the chord The WHOLE of the first bracket will ALWAYS cancel out with parts of the second bracket The numerator left will ALWAYS have a common factor of You can ALWAYS cancel So this is the gradient of the chord

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So now we take the LIMIT as Gets closer and closer to ZERO TO CONCLUDE

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Differentiating using First Principals. To find the gradient of a tangent at the point A Equation of the tangent takes the form y = mx +c m is the gradient.

Differentiating using First Principals. To find the gradient of a tangent at the point A Equation of the tangent takes the form y = mx +c m is the gradient.

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