Study Notes: Chain Rule, Quotient Rule, and Tangent Line Problems

Chain Rule Essentials

• The chain rule is about differentiating a composition of functions. If you have a composition g(h(x)), then the derivative is
  (g\circ h)'(x) = g'(h(x))\cdot h'(x).
• Think of outer function first (what you evaluate last) and inner function second (what you differentiate first). Then multiply the derivatives.
• For more than two layers, apply the rule step by step. If you have f(x) = F(G(H(x))), then
  f'(x) = F'(G(H(x)))\cdot G'(H(x))\cdot H'(x).
• In practice, you identify:
  • Outer function g and its derivative g',
  • Inner function h and its derivative h',
  • Then multiply g'(h(x)) by h'(x).
• Example worked through in class: f(x) = (x^3+5)^2 is a chain of two functions with g(u)=u^2 and h(x)=x^3+5.
  • h'(x) = 3x^2, g'(u) = 2u, so
    f'(x) = g'(h(x))\cdot h'(x) = 2(x^3+5)\cdot(3x^2) = 6x^2(x^3+5).
• Another common chain-rule form: if you see a power outside of a composite, you can use the power rule on the outer function and then differentiate the inside.
• Practical note: there are multiple ways to differentiate the same function (e.g., using the chain rule vs expanding or using product/quotient rules). The chain-rule approach is often the most efficient, especially for high powers or nested composites.

Quotient Rule and Negative Power Reformulations

• A quotient like (\dfrac{1}{1+x^2}) can be written as a negative power:
  \frac{1}{1+x^2} = (1+x^2)^{-1}.
• If you have a quotient, rewriting it as a negative power often lets you use the chain rule directly instead of the quotient rule.
• Example: differentiate f(x) = \left(\frac{1}{1+x^2}\right)^{100} = (1+x^2)^{-100}.
  • Let (g(x)=(1+x^2)^{-1}) so (f(x)=g(x)^{100}). Then
    f'(x) = 100\,g(x)^{99}\cdot g'(x).
  • Now compute (g'(x) = -(1+x^2)^{-2}\cdot (2x) = -\frac{2x}{(1+x^2)^2}).
  • Therefore
    f'(x) = 100\,(1+x^2)^{-99}\cdot\left(-\frac{2x}{(1+x^2)^2}\right) = -\frac{200x}{(1+x^2)^{101}}.
• This result matches what you’d get by the quotient rule; the reformulation is just another route.
• Quick takeaway: if the denominator has no x-term, quotients can often be treated by rewriting as a negative power and applying the chain rule instead of the quotient rule.
• If you forget the quotient rule, a good habit is to rewrite a quotient as a product: let f(x)=g(x)/h(x) = g(x)\,h(x)^{-1} and differentiate with the product rule plus chain rule.

Trigonometric Derivatives: Quick Recap

• Derivatives of the six basic trig functions (recall or derive as needed):
  • (\dfrac{d}{dx}\sin x = \cos x),
  • (\dfrac{d}{dx}\cos x = -\sin x),
  • (\dfrac{d}{dx}\tan x = \sec^2 x).
• Reciprocal trig functions (often you derive with the quotient rule if you don’t memorize):
  • (\dfrac{d}{dx}\sec x = \sec x\tan x),
  • (\dfrac{d}{dx}\csc x = -\csc x\cot x),
  • (\dfrac{d}{dx}\cot x = -\csc^2 x).
• When a function has a composition, identify the outer function and its derivative, then multiply by the derivative of the inside function (chain rule) or apply the product/quotient rules as needed.

Tangent Line Practice: y = \sqrt{\sin x} at x = \dfrac{\pi}{6}

• Given f(x) = \sqrt{\sin x}, the tangent line at x1 = (\pi/6) uses:
  • Point: (x1 = \dfrac{\pi}{6}), (y1 = f(x_1) = \sqrt{\sin(\pi/6)}).
  • Slope: m = f'(x_1).
• Compute y1:
  • (\sin(\pi/6) = \dfrac{1}{2}), so
    y_1 = \sqrt{\dfrac{1}{2}} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}.
• Derivative via chain rule:
  • Outer function is the square root: derivative is (\dfrac{1}{2\sqrt{\sin x}}).
  • Inside derivative is (\dfrac{d}{dx}\sin x = \cos x).
  • Therefore
    f'(x) = \dfrac{\cos x}{2\sqrt{\sin x}}.
• Evaluate at x1 = (\pi/6):
  • (\cos(\pi/6) = \dfrac{\sqrt{3}}{2}), (\sin(\pi/6) = \dfrac{1}{2}), (\sqrt{\sin(\pi/6)} = \dfrac{1}{\sqrt{2}}).
  • So
    m = f'(\tfrac{\pi}{6}) = \dfrac{\sqrt{3}/2}{2\cdot(1/\sqrt{2})} = \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6}}{4}.
• Tangent line equation (point-slope form):
  • With x1 = (\pi/6) and y1 = (1/\sqrt{2}),
    y - \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{6}}{4}\biggl(x - \dfrac{\pi}{6}\biggr).
  • Solving for y gives
    y = \dfrac{\sqrt{6}}{4}x + \left( \dfrac{1}{\sqrt{2}} - \dfrac{\sqrt{6}}{4}\cdot \dfrac{\pi}{6} \right).
  • A simplified equivalent form is
    y = \dfrac{\sqrt{6}}{4}x + \left( \dfrac{1}{\sqrt{2}} - \dfrac{\sqrt{6}\,\pi}{24} \right).
• Important note on forms: WebAssign or exams may require the line in the explicit form y = mx + b; either equivalent expression is acceptable as long as the line is exact. If you simplify (or leave in a form that matches the prompt), check that you’ve included the full equation.

Additional 1-Variable Chain Rule Examples

• Example: f(x) = \sin(\sqrt{x})
  • Treat as a composition with g(u) = \sin u and h(x) = \sqrt{x} = x^{1/2}.
  • f'(x) = g'(h(x))\cdot h'(x) = \cos(\sqrt{x})\cdot \dfrac{1}{2}\,x^{-1/2} = \dfrac{\cos(\sqrt{x})}{2\sqrt{x}}.
• Example: f(x) = \tan(x) and f(x) = x\sin\left(\dfrac{1}{x}\right) (product-rule example)
  • For f(x) = x\sin(1/x): using the product rule with u=x and v=\sin(1/x),
    f'(x) = u'v + uv' = 1\cdot\sin(1/x) + x\cdot\cos(1/x)\cdot(-1/x^2) = \sin(1/x) - \dfrac{\cos(1/x)}{x}.
• Example: f(x) = \sin(1/x) is a composition that can be treated with the chain rule using u = 1/x.
• Recap: tangent line to a curve y=f(x) requires the slope m = f'(x1) and the point (x1, y1) on the curve. Then use y - y1 = m(x - x1).

Multi-layer Chain Rule: Three or More Layers

• When you have several layers, apply the chain rule successively from outside in:
  • For f(x) = F(G(H(x))) with outer F, middle G, inner H,
    f'(x) = F'(G(H(x)))\cdot G'(H(x))\cdot H'(x).
• Example: f(x) = \sqrt{\tan(x^2)} = (\tan(x^2))^{1/2}.
  • Outer function: y = u^{1/2}, derivative is (1/2)u^{-1/2} times derivative of the inside.
  • Middle function: u = \tan(v), derivative is sec^2(v) times derivative of v.
  • Inner function: v = x^2, derivative is 2x.
  • Combine:
    f'(x) = \frac{1}{2}\tan(x^2)^{-1/2} \cdot \sec^2(x^2) \cdot 2x = \frac{ x\sec^2(x^2) }{ \sqrt{\tan(x^2)} }.
• This illustrates the repeated application of the chain rule: outermost, then one inner layer, and so on.

Product Rule and Higher-Order Derivatives: Quick Notes

• Product rule: if f(x) = u(x)\,v(x), then
  f'(x) = u'(x)\,v(x) + u(x)\,v'(x).
• Example (product with chain rule inside): f(x) = x\sin\left(\frac{1}{x}\right)
  • f'(x) = 1\cdot\sin(1/x) + x\cdot\cos(1/x)\cdot(-1/x^2) = \sin(1/x) - \frac{\cos(1/x)}{x}.
• Second derivatives of such products can become lengthy; practice shows that some cancellations simplify the expression (as with the previous example).

Tangent Line Problem: Tangent to y = \tan\left(\frac{\pi}{4} x^2\right) at (1,1)

• Define f(x) = tan((\pi/4)x^2). Then:
  • f(1) = tan((\pi/4)\cdot 1^2) = tan(\pi/4) = 1, so the point is (1,1).
  • f'(x) = sec^2((\pi/4)x^2) \cdot (\pi/4)\cdot 2x = (\pi/2) x \cdot \sec^2\left((\pi/4)x^2\right).
  • Slope at x=1:
    m = f'(1) = (\pi/2)\cdot 1 \cdot \sec^2(\pi/4) = (\pi/2)\cdot 2 = \pi.
• Tangent line equation (point-slope):
  y - 1 = \pi\,(x - 1) \quad\Rightarrow\quad y = \pi x + (1 - \pi). $$
• Alternative confirmation of slope: since (\sec(\pi/4) = \sqrt{2}), (\sec^2(\pi/4) = 2); thus (m = (\pi/2) \cdot 2 = \pi).
• Quick numerical check: (\cos(\pi/4) = \sqrt{2}/2) and (\sin(\pi/4) = \sqrt{2}/2) are consistent with the trig identities used in evaluating the slope.

Practical Tips and Exam Readiness

• When in doubt, rewrite a quotient as a product with a negative exponent or as a small nested composition; this often simplifies the differentiation path.

• For complicated chain-rule problems, write down the outermost function first, differentiate it, then multiply by the derivative of the inside, repeating as needed.

• Remember to substitute exact values for standard angles (e.g., pi/6, pi/4) and simplify trig values carefully; check whether further simplification (like rationalizing denominators) improves readability.

• If your final answer can be written in multiple equivalent forms, pick the form that matches the prompt or that is easiest to verify; partial credit may depend on showing the full setup.

• When using software (WebAssign, etc.), be mindful of exponent notation and parentheses; avoid ambiguous exponent expressions to prevent parsing errors.

• Always show the full equation for a tangent line: y - y1 = m(x - x1) with the computed m, x1, y1 clearly substituted.

• Connections to foundational principles:

  • Chain rule builds on the power rule and the derivative of elementary functions (sin, cos, exp, etc.).
  • Product and quotient rules underpin many real-world modeling expressions where quantities multiply or divide.
  • Trigonometric derivatives connect to geometric meanings and unit circle values; familiarity with common angles (e.g., 0, pi/6, pi/4, pi/3, pi/2) is essential.
  • Conceptually, differentiation rules are ways to handle rates of change through composed processes, which mirrors how complex systems are built from simpler subsystems.