2026 Fall Section 3.5 The Chain Rule Part 1
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Overview
The lesson develops the chain rule from composite-function notation, then applies it to powers, roots, trigonometric functions, exponentials, products, and quotients. It connects the rules to instantaneous-rate problems involving transit use, bacteria, tiger growth, and glass recycling, emphasizing unit interpretation and choosing the simplest valid differentiation method.
Key takeaways
- For a composite function f(g(x)), differentiate the outer function while retaining its inner argument, then multiply by g'(x).
- The extended power rule applies to any differentiable inner expression: d/dx[g(x)]ⁿ=n[g(x)]ⁿ⁻¹g'(x).
- For b^{g(x)} with b>0, the derivative is b^{g(x)}ln(b)g'(x); the ln(b) factor is required whenever the base is not e.
- Before using the quotient rule, check whether algebraic simplification or a negative-exponent rewrite makes the expression easier to differentiate.
- In related-rate problems, use the intermediate variable to form a chain such as dL/dt=(dL/dW)(dW/dt), then evaluate at the requested time and report units.
- For the glass-recycling model g(p)=√(0.5p²+7), p(t)=4.4+0.1t², the rate at four years is 0.48 tons per year, not 48.
Chapters
- Reviews secant-line average rate of change and the difference quotient, (f(a+h)-f(a))/h.
- Defines the tangent slope using a limit and gives the point-slope form for a tangent line.
- Recaps power, constant, exponential, trigonometric, sum, product, and quotient rules.
- Uses f(x)=x³ and g(x)=1+x² to find f(g(x))=(1+x²)³ and g(f(x))=1+x⁶.
- Shows that function composition is generally not commutative: f(g(x)) and g(f(x)) differ.
- Notes that composing a function with its inverse returns x, where the inverse relationship applies.
- States d/dx[f(g(x))]=f'(g(x))g'(x), requiring the derivative of both the outer and inner functions.
- Expresses the rule as dy/dx=(dy/du)(du/dx) when y depends on u and u depends on x.
- Clarifies that Leibniz notation suggests cancellation, but the chain rule is a derivative relationship rather than ordinary fraction cancellation.
- Differentiates sin(2x) as 2cos(2x), retaining the original argument in the cosine.
- Differentiates sin(x²) as 2x cos(x²), multiplying by the derivative of the inner function x².
- Compares Leibniz notation with composition for y=(10x³+2)⁵, obtaining 150x²(10x³+2)⁴.
- Generalizes the power rule to [g(x)]ⁿ, giving n[g(x)]ⁿ⁻¹g'(x); for example, √(10x+5) differentiates to 5/√(10x+5).
- Keeps a constant coefficient outside the differentiation, as in 5(3x-2)⁴, whose derivative is 60(3x-2)³.
- Applies the product rule to 5x(3x-2)⁴ and factors the result as 5(3x-2)³(15x-2).
- Rewrites 1/(7x²-x)⁴ as (7x²-x)⁻⁴, then applies the extended power rule instead of the quotient rule.
- For x/(2x-1)⁴, combines the quotient and chain rules to obtain (-6x-1)/(2x-1)⁵.
- Emphasizes using the quotient rule only when needed, since rewriting with negative exponents can simplify differentiation.
- Simplifies (x²-16)/(x-4) to x+4 before applying the extended power rule to its sixth power.
- Differentiates cos(3t⁴) by retaining the argument and multiplying by its derivative, 12t³.
- For ((2x+3)/(x-8))⁵, combines the extended power and quotient rules; the inner quotient derivative is -19/(x-8)².
- Uses d/dx[e^{g(x)}]=e^{g(x)}g'(x), keeping the exponential unchanged before differentiating its exponent.
- Finds d/dt[e^{t²}]=2te^{t²} and d/dt[e^{2t}]=2e^{2t}.
- Reinforces the procedure: write the original exponential, differentiate the exponent, and multiply by that derivative.
- Derives d/dx[bˣ]=bˣ ln(b) by rewriting bˣ as e^{x ln(b)}.
- Extends the formula to b^{g(x)}: multiply by both ln(b) and g'(x).
- For 2^{7x}, obtains 7 ln(2)·2^{7x}, distinguishing the base factor from the exponent's derivative.
- Models ridership as n(t)=6t²+500t+8,000 people and differentiates to n'(t)=12t+500.
- Evaluates n'(8)=596 and interprets the positive value as ridership increasing by 596 people per week.
- Uses one specified time point to identify the question as an instantaneous-rate problem and includes units.
- Uses P(t)=(24t+10)/(t²+1) million bacteria; substitution gives P(1)=17 million.
- Applies the quotient rule to find P'(t)=(-24t²-20t+24)/(t²+1)².
- Evaluates P'(1)=-5 million bacteria per hour, indicating the population is decreasing one hour after toxin exposure.
- Combines L=25W²·⁶ millimeters with the age model W=3+0.21t kilograms; at 100 days, W=24 kg and L is about 969.369 mm.
- Uses the intermediate variable W to compute dL/dt=(dL/dW)(dW/dt)=13.65W¹·⁶.
- Evaluates the growth rate at W=24 as approximately 22.0532 millimeters per day and compares this method with composing L(t).
- Models recycled glass as g(p)=√(0.5p²+7) tons and population as p(t)=4.4+0.1t² thousand; at t=4, p=6 and g=5 tons.
- Applies the chain rule as dg/dt=(dg/dp)(dp/dt), using population as the intermediate variable rather than differentiating a more complicated composite directly.
- The stated formulas give dg/dt=0.48 tons per year at t=4; the lecture's spoken final value of 48 is inconsistent with its own arithmetic, 4.8/10=0.48.
Transcript excerpt
0:09 Uh good morning class. Here's section 3.5 the chain rule. So we covered the derivative and rates of change the derivative as a function basic differentiation formulas the product and quotient rule and now the chain rule. So if you have two points uh on a curve you create a line a blue line called a secant line. The average rate of change of wall with respect to x and there are various ways of writing it. This specific format f of a plus h minus f of a over h normally is referred to as the difference quotient. If you move this secant line in a manner that it
0:58 gets closer closer to this blue this red line which is the tangent line then we can calculate the slope of the tangent line at a frime of a is limit of f of x - f of a x - a as x approaches a or the limit of f of a plus h with this 1 - f of a over h as h approaches. To find the equation of a tangent line, we go with the point slope. y - f of a frime of a * x - a. Just stick with the point slope. Now the definition for a differentiation uh function dy dx is limit of delta y over delta x as delta x approaches zero or frime of x is limit of f of x plus h minus f of x / h. These are various formulas we've seen. Uh starting with the
1:50 power rule, the derivative of x^ 1 is n x^ n minus one. The derivative of a constant is zero. The derative of kx is k. e ^ x is a very interesting function. The derivative and the function are the same. The
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Professor Ara Araeipour.