### Rules of differentiation: Rules of computation for the derivative

### The chain rule for differentiation

Let #f# and #g# be functions. The *composition* of #f# and #g# is the function #f\circ g# given by #f\circ g(x) = f(g(x))#. The following rule provides a method for calculating the derivative of a composite function.

Chain Rule for Differentiation

The derivative of #f\circ g# is #(f'\circ g)\cdot g'#. In other words,

\[\dfrac{\dd}{\dd x}\left(f( g(x))\right) = f'(g(x))\cdot g'(x)\tiny.\]

We will not prove this rule.

The rule is intuitively rather clear: if #g(x)# has a derivatie equal to #2# in #x=a# and if #f(u)# has a derivative with value #3# in #g(a)#, then a small increase #\Delta a# of #a# results in an increase of #2 \Delta a# in #g(a)# and hence, #f# applied to #u+2 \Delta a# results in an increase of about #3 \cdot 2 \Delta a=6 \Delta a#.

When working with function rules, it may be convenient to use the following notation, introduced *earlier*:\[\frac{\dd }{\dd x}f(g(x))=\left.\frac{\dd }{\dd x}f(x)\right|_{g(x)}\cdot \frac{\dd }{\dd x}g(x)\]Here #\left.\frac{\dd }{\dd x}f(x)\right|_{a}# stands for #f'(a)#, the derivative of #f# at the point #a#. This notation is particularly useful if #f(x)# is given by a function rule: suppose #f(x)=x^2+1# and #g(x)=\frac{3}{x}#. Then #f(g(x))= \left(\frac{3}{x}\right)^2+1# the rule reads:\[\frac{\dd}{\dd x}\left(\left(\frac{3}{x}\right)^2+1\right)=\left.\frac{\dd }{\dd x}(x^2+1)\right|_{\frac{3}{x}}\cdot \frac{\dd }{\dd x}\frac{3}{x}\tiny.\]

The

*chain rule*gives that the derivative of #f\circ g# is equal to \[\begin{array}{rcl}f'(g(x))\cdot g'(x) &=& (2g(x) + 6)\cdot 4\cdot x^3\\&&\phantom{uvwxyz}\color{blue}{f'(x) = 2x+6 \text{ and }g'(x) = 4\cdot x^3 }\\&=& (2(x^4-3)+6)\cdot 4x^{3}\\&&\phantom{uvwxyz}\color{blue}{g(x) = x^4-3\text{ substituted } }\\ &=& 8\cdot x^7\\&&\phantom{uvwxyz}\color{blue}{\text{ simplified } }\end{array}\]

It is also possible to calculate the composition of #x^2+6\cdot x+4# and #x^4-3# first: #f(g(x)) = x^8-5#. The derivative then follows from

*The derivative of a polynomial*.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.