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The derivative of a function is another function which represents the slope or gradient of that function. It can also be used to represent the rate of change of one variable with respect to another - for example acceleration, the derivative of velocity, is the rate of change of velocity with respect to time. There are a number of ways of representing the derivative of an equation. Usually these depend on how you represented the function in the first place. \(\frac{d}{dx}\) is the way of representing the differentiation operative i.e. if you see \( \frac{d}{dx} ( \text{Some function)}\) then you should differentiate the function you see in the brackets. The following list shows the function on the left, and the corresponding way of writing its derivative on the right. \[\frac{d}{dx} f(x) = f'(x)\] \[\frac{d}{dx}( y )= \frac{dy}{dx} \text { or y' } \] If you are differentiating with respect to a variable other than \( x\) ,let's say \( t\) , do not use\( \frac{d}{dx}\) but \( \frac{d}{dt}\) .

Differentiation by first principles means using the idea that the deriviative represents the gradient of the curve, rather than rules and tricks, to calculating the deriviative of a function. \[ f'(x) =\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x} \] Think of the simple formula often used to calculate the slope of a line between two points: \[Slope = \frac{Rise}{Run}\] Differentiation by first principles applies this formula in the limit where \(h\), the 'Run' becomes very small - as \(h\) approaches zero, the slope between \(x\) and \(x+h\) approaches the the deriviative of the function at \(x\), or the exact slope at \(x\). Imagine moving B( in the diagram below) along the curve towards A. As B approaches A, the chord AB becomes closer and closer to the tangent at A. You may see other sources using \(\delta x\) rather than \(h\).

Example 1: Find an expression for the gradient of the function \[f(x) = 3x^2\] by differentiating from first principles. Answer: \[f'(x) =\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x}\] First, we must calculate \() f(x+h) \): \[f(x+h) = 3(x+h)^2 = 3(x^2 + 2xh + h^2)=3x^2 +6xh+3h^2\] Subtracting f(x) and subbing into the formula gives \[\frac{3x^2 +6xh+3h^2-3x^2}{(x+h)-x}= \frac{6xh +3h^2}{h} = 6x + 3h\] Now consider what happens as hh h approaches zero: The first term will not be affected and the second term will go to zero. Therefore, the derivative can be given by: \[f'(x) = 6x\]

Example 2: Find an expression for the gradient of the function \[f(x) = x^3\] by differentiating from first principles. Answer: \[ f'(x) =\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x} \] First, we must calculate \(f(x+h)\): \[f(x+h) = (x+h)^3 = x^3 +3x^2h + 3xh^2 + h^3\] Subtracting \(f(x)\) and subbing into the formula gives \[\frac{x^3 +3x^2h + 3xh^2 + h^3-x^3}{(x+h)-x}= \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2\] Now consider what happens as \(h\) approaches zero: The first term will not be affected and the second and third terms will go to zero. Therefore, the derivative can be given by: \[f'(x) = 3x^2\]

VCE biology Units one and two: Unit two - Cell cycle, asexual and sexual reproduction, meiosis, cell growth and differentiation

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