The Chain Rule
- Page ID
- 625
Our goal is to differentiate functions such as
\[ y = (3x + 1)^{10} \]
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 \[\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \] |
In our example we have
\[ y = u^{10} \]
and
\[ u = 3x + 1 \]
so that
\[\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \]
= (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 \[ f(x) = (x^4 - 3x^3 + x)^5 \] Solution Here \[ f(u) = u^5 \] and \[ u(x) = x^4 - 3x^3 + x \] 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 \[ f(x) = (x^3 - x + 1)^{20} \] Solution Here \[ f(u) = u^{20} \] and \[ u(x) = x^3 - x + 1 \] So that the derivative is \[ (20u^{19})(3x^2 - 1) = \left[20(x^3 - x + 1)^{19}\right](3x^2 - 1) \] |
Example 4 |
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Find f '(x) if \[ f(x) = (1 - x)^9 (1-x^2)^4 \] Solution Here 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 \[ f(x)= \dfrac{ (x^3 + 4x - 3)^7}{ (2x - 1)^3} \] Solution Here we need both the quotient and the chain rule. \[ f'(x) = \dfrac{(2x - 1)^3\left[(x^3 + 4x - 3)^7\right]' - (x^3 + 4x - 3)^7 \left[(2x - 1)^3\right]'}{(2x - 1)^6} \] 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 \[ f](x) = \dfrac{7(2x - 1)^3(x^3 + 4x - 3)^6(3x^2 + 4) + 6(x^3 + 4x - 3)^7 (2x - 1)^2}{ (2x - 1)^6 } \] |
\[ f(x)= \dfrac{x^2(5 - x^3)^4 }{3 - x} \]