The Chain Rule
( \newcommand{\kernel}{\mathrm{null}\,}\)
Our goal is to differentiate functions such as
y = (3x + 1)^{10} \nonumber
The last thing that we would want to do is FOIL this out ten times. We now look for a better way.
Definition: The Chain Rule |
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If y = y(u) is a function of u, and u = u(x) is a function of x then \dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \nonumber |
In our example we have
y = u^{10} \nonumber
and
u = 3x + 1 \nonumber
so that
\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \nonumber
= (10u9)(3) = 30(3x+1)9
Proof: Chain Rule |
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Recall an alternate definition of the derivative: |
Example 2 |
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Find f '(x) if f(x) = (x^4 - 3x^3 + x)^5 \nonumber SolutionHere f(u) = u^5 \nonumber and u(x) = x^4 - 3x^3 + x \nonumber So that the derivative is (5u4)(4x3 - 9x2 + 1) = [5(x4 - 3x3 + x)4](4x3 - 9x2 + 1) |
Example 3 |
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Find f '(x) if f(x) = (x^3 - x + 1)^{20} \nonumber SolutionHere f(u) = u^{20} \nonumber and u(x) = x^3 - x + 1 \nonumber So that the derivative is (20u^{19})(3x^2 - 1) = \left[20(x^3 - x + 1)^{19}\right](3x^2 - 1) \nonumber |
Example 4 |
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Find f '(x) if f(x) = (1 - x)^9 (1-x^2)^4 \nonumber SolutionHere we need both the product and the chain rule. First the product rule f '(x) = [(1 - x)9][(1 - x2)4] ' + [(1 - x)9]' [(1 - x2)4] Now compute [(1 - x2)4]' = [4(1 - x2)3](-2x) and [(1 - x)9]' = [9(1 - x)8](-1) |
Example 5 |
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Find f '(x) if f(x)= \dfrac{ (x^3 + 4x - 3)^7}{ (2x - 1)^3} \nonumber SolutionHere we need both the quotient and the chain rule. f'(x) = \dfrac{(2x - 1)^3\left[(x^3 + 4x - 3)^7\right]' - (x^3 + 4x - 3)^7 \left[(2x - 1)^3\right]'}{(2x - 1)^6} \nonumber We first compute [(x3 + 4x - 3)7]' = [7(x3 + 4x - 3)6](3x2 + 4) and [(2x - 1)3]' = [3(2x - 1)2](2) Putting this all together gives f](x) = \dfrac{7(2x - 1)^3(x^3 + 4x - 3)^6(3x^2 + 4) + 6(x^3 + 4x - 3)^7 (2x - 1)^2}{ (2x - 1)^6 } \nonumber |