Chain Rule

The Chain Rule tells us how to differentiate composite functions. As a refresher, composite functions are nested functions or functions within functions. They are commonly expressed as:

\(f(g(x))\)

Chain Rule can be algebraically expressed in Prime Notation as:

\(h'(x) = [f(g(x))]' = \textcolor{red}{f'(g(x))}\cdot \textcolor{blue}{g'(x)}\)

  • \(h'(x)\) represents the derivative of the Chain function
  • \(\textcolor{red}{f'(g(x))}\) represents the inner function nested within the derivative of the outer function
  • \(\textcolor{blue}{g'(x)}\) represents the derivative of the inner function

In short, Chain Rule takes the derivative of the outer function while leaving the inner function, multiplied by the derivative of the inner function.


Example

Find the derivative of \(p(u(x))\) where \(p(x) = \sqrt{x}\) and \(u(x) = 4x^2 -3x\).

First, we can rewrite \(p(u)\). We can do this by substituting \(u\) for \(x\) and express it in exponent form:

\(p(u) = \sqrt{u}\)

\(p(u) = u^{\frac{1}{2}}\)

Next, we can determine \(p'(u)\):

\(p'(u) = \cfrac{1}{2}u^{-\frac{1}{2}}\)

\(p'(u) = \cfrac{1}{2u^{\frac{1}{2}}}\)

Then, we can determine \(p'(u(x))\) by substituting \(u(x)\) for \(u\):

\(\textcolor{red}{p'(u(x)) = \cfrac{1}{2(4x^2 -3x)^{\frac{1}{2}}}}\)

After, we can determine \(u'(x)\):

\(u'(x) = 2(4x) -1(3x)\)

\(\textcolor{blue}{u'(x) = 8x -3}\)

Finally, we can put everything together to determine the derivative of the composite function:

\([p(u(x))]' = \textcolor{red}{p'(u(x))}\cdot \textcolor{blue}{u'(x)}\)

\([p(u(x))]' = \left(\textcolor{red}{\cfrac{1}{2(4x^2 -3x)^{\frac{1}{2}}}}\right)(\textcolor{blue}{8x -3})\)

\([p(u(x))]' = \cfrac{8x -3}{2(4x^2 -3x)^{\frac{1}{2}}}\)

Therefore, we can determine that \([p(u(x))]' = \cfrac{8x -3}{2(4x^2 -3x)^{\frac{1}{2}}}\).


Find the derivative of \(b(p(x))\) where \(b(x) = \cfrac{1}{x}\) and \(p(x) = x^3 + 5x\).

Example

If \(y = -\sqrt{u}\) and \(u = 4x^3-3x^2+1\), determine \(\cfrac{dy}{dx}\) at \(x=0\).

First, we can rewrite \(y(u)\) as such:

\(y = -u^{\frac{1}{2}}\)

Next, we can determine \(y'(u)\):

\(y' = -\cfrac{1}{2}(u^{-\frac{1}{2}})\)

\(y' = -\cfrac{1}{2u^{\frac{1}{2}}}\)

Then, we can determine \(y'(u(x))\) by substituting \(u(x)\) for \(u\):

\(\textcolor{red}{y'(u(x)) = -\cfrac{1}{2(4x^3-3x^2+1)^{\frac{1}{2}}}}\)

After, we can determine \(u'(x)\):

\(u' = 3(4x^2)-2(3x)\)

\(\textcolor{blue}{u' = 12x^2-6x}\)

Finally, we can put everything together and subsitute \(1\) for \(x\):

\([y(u(x))]' = \textcolor{red}{y'(u(x))}\cdot \textcolor{blue}{u'(x)}\)

\([y(u(x))]' = \left(\textcolor{red}{-\cfrac{1}{2(4x^3-3x^2+1)^{\frac{1}{2}}}}\right)(\textcolor{blue}{12x^2-6x})\)

\([y(u(0))]' = \left[-\cfrac{1}{2(4(0)^3-3(0)^2+1)^{\frac{1}{2}}}\right][12(0)^2-6(0)]\)

\([y(u(0))]' = \left[-\cfrac{1}{2(1)^{\frac{1}{2}}}\right][0]\)

\([y(u(0))]' = \left(-\cfrac{1}{2}\right)(0)\)

\([y(u(0))]' = 0\)

Therefore, we can determine that \(\cfrac{dy}{dx}\) at \(x=0\) is \(0\).


If \(a = b(2-b^2)\) and \(b = \cfrac{1}{c}\), determine \(\cfrac{da}{dc}\) at \(c=2\).

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