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:
Chain Rule can be algebraically expressed in Prime Notation as:
In short, Chain Rule takes the derivative of the outer function while leaving the inner function, multiplied by the derivative of the inner function.
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 \(\boldsymbol{[p(u(x))]' = \cfrac{8x -3}{2(4x^2 -3x)^{\frac{1}{2}}}}\).
First, we can find \(b(p)\). We can do this by substituting \(p\) for \(x\):
\(b(p) = \cfrac{1}{p}\)
Next, we can determine \(b'(p)\):
\(b(p) = -p^{-1}\)
\(b'(p) = -1(p^{-2})\)
\(b'(p) = -\cfrac{1}{p^{2}}\)
Then, we can determine \(b'(p(x))\) by substituting \(p(x)\) for \(p\):
\(\textcolor{red}{b'(p(x)) = -\cfrac{1}{(x^3 + 5x)^{2}}}\)
After, we can determine \(p'(x)\):
\(p'(x) = 3(x^2) + 5\)
\(\textcolor{blue}{p'(x) = 3x^2 + 5}\)
Finally, we can put everything together to determine the derivative of the composite function:
\([b(p(x))]' = \textcolor{red}{b'(p(x))}\cdot \textcolor{blue}{p'(x)}\)
\([b(p(x))]' = \left(\textcolor{red}{-\cfrac{1}{(x^3 + 5x)^{2}}}\right)(\textcolor{blue}{3x^2 + 5})\)
\([b(p(x))]' = -\cfrac{3x^2 + 5}{(x^3 + 5x)^{2}}\)
Therefore, we can determine that \(\boldsymbol{[b(p(x))]' = -\cfrac{3x^2 + 5}{(x^3 + 5x)^{2}}}\).
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 \(y'(u)\):
\(y' = -\cfrac{1}{2u^{\frac{1}{2}}}\)
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 \(\boldsymbol{0}\).
First, we can determine \(a'(b)\):
\(a(b) = 2b-b^3\)
\(a'(b) = 1(2)-3(b^2)\)
\(a'(b) = 2-3b^2\)
Next, we can determine \(a'(b(c))\) by substituting \(b(c)\) for \(b\):
\(\textcolor{red}{a'(b(c)) = 2-3 \left(\cfrac{1}{c}\right)^2}\)
Then, we can determine \(b'(c)\):
\(b = c^{-1}\)
\(b'(c) = -1(c^{-2})\)
\(\textcolor{blue}{b'(c) = -\cfrac{1}{c^{2}}}\)
Finally, we can put everything together and subsitute \(b\) and \(c\) with their respective values:
\([a(b(c))]' = \textcolor{red}{a'(b(c))}\cdot \textcolor{blue}{b'(c)}\)
\([a(b(c))]' = \left[\textcolor{red}{2-3\left(\cfrac{1}{c}\right)^2}\right] \left[\textcolor{blue}{-\cfrac{1}{c^{2}}}\right]\)
\([a(b(2))]' = \left[2-3 \left(\cfrac{1}{2}\right)^2\right] \left[-\cfrac{1}{(2)^{2}}\right]\)
\([a(b(2))]' = \left[2-3 \left(\cfrac{1}{4}\right)\right] \left[-\cfrac{1}{4}\right]\)
\([a(b(2))]' = \left(\cfrac{8}{4}-\cfrac{3}{4}\right) \left(-\cfrac{1}{4}\right)\)
\([a(b(2))]' = \left(\cfrac{5}{4}\right) \left(-\cfrac{1}{4}\right)\)
\([a(b(2))]' = -\cfrac{5}{16}\)
Therefore, we can determine that \(\cfrac{da}{dc}\) at \(c=2\) is \(\boldsymbol{-\cfrac{5}{16}}\).