### Differentiation: Calculating derivatives and tangent lines

### Chain rule for differentiation

With the chain rule we can determine the derivative of a *composite function* using the derivatives of the function from which it is composed.

Chain rule

For two functions #\blue{f(x)}# and #\green{g(x)}# holds \[

(\blue f \circ \green{g})'(x)=\blue f'(\green{g(x)}) \cdot \green{g'(x)}\]

**Example**

\[\frac{\dd}{\dd x} \blue ( \green{x^2-5x}\blue{)^4}=4 \cdot (\green{x^2-5x})^3 \cdot (2x-5)

\]

#\frac{\dd}{\dd x}\left (\left(3\cdot x^3+4\right)^{{{1}\over{3}}}\right)=# \({{3\cdot x^2}\over{\left(3\cdot x^3+4\right)^{{{2}\over{3}}}}}\)

We can calculate #\dfrac{\dd}{\dd x}\left(\left(3\cdot x^3+4\right)^{{{1}\over{3}}}\right)# by using the chain rule. Write #(f\circ g)(x)=\left(3\cdot x^3+4\right)^{{{1}\over{3}}}# with #f(x)=x^{{{1}\over{3}}}# and #g(x)=3\cdot x^3+4#. Now e can apply the chain rule which states: #(f\circ g)'(x)=f'(g(x)) \cdot g'(x)#.

\[\begin{array}{rcl}\displaystyle \dfrac{\dd}{\dd x} \left(\left(3\cdot x^3+4\right)^{{{1}\over{3}}}\right) &=& \displaystyle {{1}\over{3\cdot g(x)^{{{2}\over{3}}}}} \cdot \frac{\dd }{\dd x}\left(g(x)\right) \\

&&\phantom{xxx}\blue{\text{chain rule applied with }f'(x)={{1}\over{3\cdot x^{{{2}\over{3}}}}}}\\

&=&\displaystyle \left({{1}\over{3\cdot \left(3\cdot x^3+4\right)^{{{2}\over{3}}}}}\right)\cdot \frac{\dd }{\dd x}\left(3\cdot x^3+4\right) \\

&&\phantom{xxx}\blue{g(x)=3\cdot x^3+4 \text{ substituted}}\\

&=&\displaystyle \left({{1}\over{3\cdot \left(3\cdot x^3+4\right)^{{{2}\over{3}}}}}\right)\cdot\left( 9\cdot x^2\right) \\

&&\phantom{xxx}\blue{\frac{\dd }{\dd x}\left(3\cdot x^3+4\right) \text{calculated}}\\

&=&\displaystyle {{3\cdot x^2}\over{\left(3\cdot x^3+4\right)^{{{2}\over{3}}}}}\\

&&\phantom{xxx}\blue{\text{simplified}}

\end{array}\]

We can calculate #\dfrac{\dd}{\dd x}\left(\left(3\cdot x^3+4\right)^{{{1}\over{3}}}\right)# by using the chain rule. Write #(f\circ g)(x)=\left(3\cdot x^3+4\right)^{{{1}\over{3}}}# with #f(x)=x^{{{1}\over{3}}}# and #g(x)=3\cdot x^3+4#. Now e can apply the chain rule which states: #(f\circ g)'(x)=f'(g(x)) \cdot g'(x)#.

\[\begin{array}{rcl}\displaystyle \dfrac{\dd}{\dd x} \left(\left(3\cdot x^3+4\right)^{{{1}\over{3}}}\right) &=& \displaystyle {{1}\over{3\cdot g(x)^{{{2}\over{3}}}}} \cdot \frac{\dd }{\dd x}\left(g(x)\right) \\

&&\phantom{xxx}\blue{\text{chain rule applied with }f'(x)={{1}\over{3\cdot x^{{{2}\over{3}}}}}}\\

&=&\displaystyle \left({{1}\over{3\cdot \left(3\cdot x^3+4\right)^{{{2}\over{3}}}}}\right)\cdot \frac{\dd }{\dd x}\left(3\cdot x^3+4\right) \\

&&\phantom{xxx}\blue{g(x)=3\cdot x^3+4 \text{ substituted}}\\

&=&\displaystyle \left({{1}\over{3\cdot \left(3\cdot x^3+4\right)^{{{2}\over{3}}}}}\right)\cdot\left( 9\cdot x^2\right) \\

&&\phantom{xxx}\blue{\frac{\dd }{\dd x}\left(3\cdot x^3+4\right) \text{calculated}}\\

&=&\displaystyle {{3\cdot x^2}\over{\left(3\cdot x^3+4\right)^{{{2}\over{3}}}}}\\

&&\phantom{xxx}\blue{\text{simplified}}

\end{array}\]

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