14.1 Chain Rule
Let \(u=g(x)\) and \(y=f(u)\). Then \(y\) is a function of \(x\) and
\[\frac {dy}{dx}=\frac {dy}{du}\,\cdot \, \frac {du}{dx}\]
Solution.
\(\text {Let}\quad u=x^3+2x\), then \(y = \sin u\).
\[\implies \frac {dy}{du}=\cos u\] \[u=x^3+2x\quad \implies \quad \frac {du}{dx}=3x^2+2\]
\begin {align*} \text {Then}\qquad \frac {dy}{dx} & =\frac {dy}{du}\cdot \frac {du}{dx}\\\\ & = (\cos u)\cdot (3x^2+2)\\\\ \therefore \quad \frac {dy}{dx} & = (3x^2+2)\cos (x^3+2x)\\ \end {align*}
Find \(\frac {dy}{dx}\) if
- 1.
- \(y=(3x^5-1)^{20}\)
- 2.
- \(y=\ln (x^2-2x)\)
- 3.
- \(y=e^{-3x^2}\)
Solution. (i). With \(u=3x^{5}-1\), \(y=u^{20}\), so \(\frac {dy}{du}=20u^{19}\) and \(\frac {du}{dx}=15x^{4}\): \[\frac {dy}{dx}=20\left (3x^{5}-1\right )^{19}\cdot 15x^{4} =300x^{4}\left (3x^{5}-1\right )^{19}.\] Expanding the twentieth power first would have been unthinkable; that is what the chain rule buys.
(ii). With \(u=x^{2}-2x\), \(y=\ln u\), so \(\frac {dy}{du}=\frac {1}{u}\): \[\frac {dy}{dx}=\frac {1}{x^{2}-2x}\cdot (2x-2)=\frac {2x-2}{x^{2}-2x} =\frac {2(x-1)}{x(x-2)} .\]
(iii). With \(u=-3x^{2}\), \(y=e^{u}\), so \(\frac {dy}{du}=e^{u}\): \[\frac {dy}{dx}=e^{-3x^{2}}\cdot (-6x)=-6x\,e^{-3x^{2}} .\]
Note 14.13. The pattern in all three is worth naming: the derivative of the outside function, evaluated at the inside, times the derivative of the inside. In shorthand, \[\frac {d}{dx}f\left (g(x)\right )=f'\left (g(x)\right )\cdot g'(x).\] The factor \(g'(x)\) is the part most often forgotten, and its absence is easy to spot — without it, \(\frac {d}{dx}e^{-3x^{2}}\) would come out as \(e^{-3x^{2}}\), whose value at \(x=0\) would say the curve is rising when in fact it has a maximum there.
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