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== Differentiating Logarithmic and Exponential functions ==

The function <math>y = e^x</math> is its own derivative: <math>\dfrac{d}{dx} e^x = e^x</math>. The constant <math>e</math> is defined such that this is true.

An exponential function with a different base can be converted into a function of the form <math>e^{kx}</math> using logarithms, e.g. <math>2^x = e^{\ln 2 x}</math>. The derivative of such an expression can be found using the chain rule: <math>\dfrac{d}{dx} 2^x = \dfrac{d}{dx} e^{\ln 2 x} = \ln 2\ e^{\ln 2 x} = 2^x \ln 2</math>.

<math>\dfrac{d}{dx} e^{f(x)} = f'(x)e^{f(x)}</math>

The derivative of a logarithm is <math>\dfrac{d}{dx} \ln x = \frac{1}{x}</math>. Applying the chain rule to this produces the result:

<math>\dfrac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}</math>

It is important to know how these rules interact with other expressions.

e.g. <math>\dfrac{d}{dx} e^{x^2 + \ln \tfrac{1}{x}} = e^{x^2 + \ln \tfrac{1}{x}} (\dfrac{d}{dx} x^2 + \ln \tfrac{1}{x}) = e^{x^2 + \ln \tfrac{1}{x}} (2x + \frac{-x^{-2}}{1/x}) = e^{x^2 + \ln \tfrac{1}{x}} (2x - x^{-1}) </math>

== Differentiating Trigonometric Functions ==

{{sidenote|The proof of these derivatives is beyond the scope of the syllabus, but we can find them using the addition formulae. |title=Proof of the trigonometric derivatives}}

The trigonometric functions have the following derivatives:

* <math>\dfrac{d}{dx} \sin x = \cos x </math>
* <math>\dfrac{d}{dx} \cos x = -\sin x </math>
* <math>\dfrac{d}{dx} \tan x = \sec^2 x </math>
* <math>\dfrac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}</math>

== The Product Rule ==

{{sidenote|Consider a rectangle where one side is of length <math>f(x)</math> and the other side is of length <math>g(x)</math>. The area of the rectangle is equivalent to the value of the product. As x increases by a small amount <math>\delta x</math>, the area changes. The new area is <math>f(x+\delta x)g(x+\delta x)</math>. Thus, the change in area is <math>f(x+\delta x)g(x+\delta x) - f(x)g(x)</math> 

which is equivalent to <math>(f(x)+\delta f)(g(x)+\delta g) - f(x)g(x)</math> 

where <math>\delta f</math> is the difference <math>f(x+\delta x)-f(x)</math> 

and <math>\delta g</math> is the difference <math>g(x+\delta x)-g(x)</math>. 

<math>\begin{align}&(f(x)+\delta f)(g(x)+\delta g) - f(x)g(x) \\ =& f(x)g(x)+f(x)\delta g + g(x)\delta f + \delta f\delta g - f(x)g(x) \\ =& f(x)\delta g + g(x)\delta f + \delta f\delta g \end{align}</math>

As <math>\delta x</math> approaches zero, the term <math>\delta f\delta g</math> becomes negligible. Thus, the change in area is <math>f(x)\delta g + g(x)\delta f</math> and because the derivative of the product is the change in area over a change in <math>x</math>, <math>\dfrac{d}{dx} f(x)g(x) = f(x)\dfrac{dg}{dx} + g(x)\dfrac{df}{dx}</math>|title=Proof of the Product Rule|width=40%}}

The product rule states that:

<math>\dfrac{d}{dx} f(x)g(x) = f(x)g'(x) + g(x)f'(x)</math>

e.g. <math>\dfrac{d}{dx} xe^x = x\dfrac{d}{dx}e^x + e^x\dfrac{d}{dx}x = xe^x + e^x = e^x(x + 1)</math>

=== The Quotient Rule ===

The quotient rule is a special case of the product rule when one of the terms in the product is a reciprocal.

e.g. Evaluate <math>\dfrac{d}{dx} \frac{2x+3}{x+4} </math>

<math>\begin{align}
\dfrac{d}{dx} \frac{2x+3}{x+4} &= \dfrac{d}{dx} (2x+3)\frac{1}{x+4}\\
&= (2x+3)\dfrac{d}{dx}\left(\frac{1}{x+4}\right) + \frac{1}{x+4}\dfrac{d}{dx}(2x+3) \\
&= -(2x+3)(x+4)^{-2} + \frac{1}{x+4}(2) \\
&= \frac{-(2x+3)}{(x+4)^2} + \frac{2}{x+4} \\
&= \frac{-(2x+3)}{(x+4)^2} + \frac{2x + 8}{(x+4)^2} \\
&= \frac{2x + 8 - 2x - 3}{(x+4)^2} \\
&= \frac{5}{(x+4)^2} 
\end{align}</math>

In general:

<math>\begin{align}
\dfrac{d}{dx} \frac{u}{v} &= u\dfrac{d}{dx}\frac{1}{v} + \frac{1}{v}\dfrac{du}{dx} \\
&= u(-v)^{-2}\dfrac{dv}{dx} + \frac{v}{v^2}\dfrac{du}{dx} \\
&= \frac{-u\tfrac{dv}{dx} + v\tfrac{du}{dx}}{v^2} \\
&= \frac{v\tfrac{du}{dx} - u\tfrac{dv}{dx}}{v^2}
\end{align}</math>

== Implicit Differentiation ==

Implicit differentiation is where we differentiate a function which is not defined explicitly, with y as the subject. To do this, it is sensible to use the chain rule.

e.g. Find an expression for <math>\dfrac{dy}{dx}</math> when <math>x^2 + y^2 = 1</math>.

<math>\begin{align}
x^2 + y^2 &= 1 \\
\dfrac{d}{dx} x^2 + y^2 &= \dfrac{d}{dx} 1 \\
2x + \dfrac{dy^2}{dx} &= 0 \\
2x + \dfrac{dy^2}{dy}\dfrac{dy}{dx} &= 0 \quad\text{Use the chain rule} \\
2x + 2y\dfrac{dy}{dx} &= 0 \\
2y\dfrac{dy}{dx} &= -2x \\
\dfrac{dy}{dx} &= -\frac{2x}{2y} = -\frac{x}{y} 
\end{align}</math>

Sometimes, we need to use the product rule too.

e.g. Find an expression for <math>\dfrac{dy}{dx}</math> when <math>x^2 + 6xy + y^2 = 1</math>.

<math>\begin{align}
x^2 + 6xy + y^2 &= 1 \\
\dfrac{d}{dx} x^2 + 6xy + y^2 &= \dfrac{d}{dx} 1 \\
2x + \dfrac{d}{dx}6xy + \dfrac{d}{dx}y^2 &= 0 \\
2x + 6x\dfrac{dy}{dx} + y\dfrac{d}{dx}6x + \dfrac{dy^2}{dy}\dfrac{dy}{dx} &= 0 \\
2x + 6x\dfrac{dy}{dx} + 6y + 2y\dfrac{dy}{dx} &= 0 \\
2x + 6y + (6x + 2y)\dfrac{dy}{dx} &= 0 \\
(6x + 2y)\dfrac{dy}{dx} &= - 2x - 6y \\
\dfrac{dy}{dx} &= \frac{-2x-6y}{6x+2y} \\
\dfrac{dy}{dx} &= \frac{-x-3y}{3x+y} 
\end{align}</math>

== Parametric Differentiation ==

A parametric function is where instead of <math>y</math> being defined by <math>x</math>, <math>y</math> and <math>x</math> are both linked to a third parameter, <math>t</math>. e.g. <math>x = 3t, y = t^2</math>

To find <math>\dfrac{dy}{dx}</math> when <math>y</math> and <math>x</math> are defined parametrically, we need to use the chain rule:

<math>\dfrac{dy}{dx} = \dfrac{dy}{dt}\times\dfrac{dt}{dx} = \dfrac{dy}{dt}\div\dfrac{dx}{dt}</math>

So for the example <math>x = 3t, y = t^2</math>, <math>\dfrac{dx}{dt} = 3</math> and <math>\dfrac{dy}{dt} = 2t</math>, thus <math>\dfrac{dy}{dx} = \dfrac{dy}{dt}\div\dfrac{dx}{dt} = 2t \div 3 = \frac{2t}{3}</math>

{{chapnav|Trigonometry|Integration}}

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