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 ({{1}\over{\left(x+2\right)^{{{3}\over{2}}}}}\right)=# \(-{{3}\over{2\cdot \left(x+2\right)^{{{5}\over{2}}}}}\)
We can calculate #\dfrac{\dd}{\dd x}\left({{1}\over{\left(x+2\right)^{{{3}\over{2}}}}}\right)# by using the chain rule. Write #(f\circ g)(x)={{1}\over{\left(x+2\right)^{{{3}\over{2}}}}}# with #f(x)=x^{{{3}\over{2}}}# and #g(x)={{1}\over{x+2}}#. Now e can apply the chain rule which states: #(f\circ g)'(x)=f'(g(x)) \cdot g'(x)#.
\[\begin{array}{rcl}\displaystyle \dfrac{\dd}{\dd x} \left({{1}\over{\left(x+2\right)^{{{3}\over{2}}}}}\right) &=& \displaystyle {{3\cdot \sqrt{g(x)}}\over{2}} \cdot \frac{\dd }{\dd x}\left(g(x)\right) \\
&&\phantom{xxx}\blue{\text{chain rule applied with }f'(x)={{3\cdot \sqrt{x}}\over{2}}}\\
&=&\displaystyle \left({{3}\over{2\cdot \sqrt{x+2}}}\right)\cdot \frac{\dd }{\dd x}\left({{1}\over{x+2}}\right) \\
&&\phantom{xxx}\blue{g(x)={{1}\over{x+2}} \text{ substituted}}\\
&=&\displaystyle \left({{3}\over{2\cdot \sqrt{x+2}}}\right)\cdot\left( -{{1}\over{\left(x+2\right)^2}}\right) \\
&&\phantom{xxx}\blue{\frac{\dd }{\dd x}\left({{1}\over{x+2}}\right) \text{calculated}}\\
&=&\displaystyle -{{3}\over{2\cdot \left(x+2\right)^{{{5}\over{2}}}}}\\
&&\phantom{xxx}\blue{\text{simplified}}
\end{array}\]
We can calculate #\dfrac{\dd}{\dd x}\left({{1}\over{\left(x+2\right)^{{{3}\over{2}}}}}\right)# by using the chain rule. Write #(f\circ g)(x)={{1}\over{\left(x+2\right)^{{{3}\over{2}}}}}# with #f(x)=x^{{{3}\over{2}}}# and #g(x)={{1}\over{x+2}}#. Now e can apply the chain rule which states: #(f\circ g)'(x)=f'(g(x)) \cdot g'(x)#.
\[\begin{array}{rcl}\displaystyle \dfrac{\dd}{\dd x} \left({{1}\over{\left(x+2\right)^{{{3}\over{2}}}}}\right) &=& \displaystyle {{3\cdot \sqrt{g(x)}}\over{2}} \cdot \frac{\dd }{\dd x}\left(g(x)\right) \\
&&\phantom{xxx}\blue{\text{chain rule applied with }f'(x)={{3\cdot \sqrt{x}}\over{2}}}\\
&=&\displaystyle \left({{3}\over{2\cdot \sqrt{x+2}}}\right)\cdot \frac{\dd }{\dd x}\left({{1}\over{x+2}}\right) \\
&&\phantom{xxx}\blue{g(x)={{1}\over{x+2}} \text{ substituted}}\\
&=&\displaystyle \left({{3}\over{2\cdot \sqrt{x+2}}}\right)\cdot\left( -{{1}\over{\left(x+2\right)^2}}\right) \\
&&\phantom{xxx}\blue{\frac{\dd }{\dd x}\left({{1}\over{x+2}}\right) \text{calculated}}\\
&=&\displaystyle -{{3}\over{2\cdot \left(x+2\right)^{{{5}\over{2}}}}}\\
&&\phantom{xxx}\blue{\text{simplified}}
\end{array}\]
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