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Added two examples and the rules for finding derivatives section
:<math>\frac {dy}{dx} = \frac {dy} {du} \times \frac {du}{dx}.</math>
:<math>\frac {dy}{dx} = \frac {dy} {du} \times \frac {du}{dx}.</math>
The chain rule can also be applied to multivariable functions. The derivative of a multivariable function is expressed as follows:
The chain rule can also be applied to multivariable functions. The derivative of a function with two variables is expressed as follows:
<math> \frac {d}{dt}(f(x(t), y(t))) = \frac{\partial f}{\partial x}\times \frac{dx}{dt} + \frac{\partial f}{\partial y}\times \frac{dy}{dt} </math>
<math> \frac {d}{dt}(f(x(t), y(t))) = \frac{\partial f}{\partial x}\times \frac{dx}{dt} + \frac{\partial f}{\partial y}\times \frac{dy}{dt} </math>
Sometimes it might be helpful to [[Change of variables|change variables]] into a more convenient form before differentiating.
Sometimes it might be helpful to [[Change of variables|change variables]] into a more convenient form before differentiating.
==Examples==
===Example with one Variable===
Suppose we wish to find the derivative of the function:
<math>
f(x) = \sin{\big( x^2 \big)}
</math>
However, we do not know how to find this but we do know the derivative of [[sine]]. Therefore, we can use the chain rule using x<sup>2</sup> as g(x) and sin(g(x)) as f(g(x)). Therefore, using u to represent g(x):
<math>
\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx} = \frac{d}{du} \big( \sin{u} \big) \times \frac{d}{dx} \big( x^2 \big)
</math>
Using standard results, these derivatives can be found and substituting back for u=x<sup>2</sup> gives:
<math>
\frac{df}{dx} = \cos{u} \times 2x = 2x \cos{\big( x^2 \big)}
</math>
===Example with two Variables===
Suppose the [[potential energy]] of a particle undergoing [[simple harmonic motion]] is described by the function V(x,y) and that both x and y are themselves functions of time. Now suppose we wish to find the rate of change potential energy with respect to time. To do this we must utilise the chain rule above. The potential for simple harmonic motion is:
<math>
V(x,y)=\frac{1}{2} k \big( x^2 + y^2 \big)
</math>
and x and y are given by: <math>
x(t)=A \sin{\omega t}
</math> and <math>
y(t)=B \sin{\omega t}
</math>
where A, B, k and ω are constants. Applying the result above gives:
<math>
\frac{dV}{dt} = \frac{\partial V}{\partial x} \times \frac{dx}{dt} + \frac{\partial V}{\partial y} \times \frac{dy}{dt} =
kx \times A\omega \cos{\omega t} + ky \times B\omega \cos{\omega t}
</math>
Substituting in for x and y produces the answer:
<math>
\frac{dV}{dt} = k\omega \big( A^2 +B^2 \big) \sin{\omega t} \cos{\omega t}
</math>
==Rules for finding derivatives==
*[[Power rule]]
*[[Constant-multiple rule]]
*[[sum rule]]
*[[Chain rule]]
*[[Product rule]]
*[[Quotient rule]]
[[Category:Mathematics]]
[[Category:Calculus]]
[[Category:Calculus]]
[[Category:Differentiation]]
[[Category:Differentiation]]