Math

(9 points each part) Find the derivative of the function. PLEASE do not simplify. Label. a) f(x) = 1/(2x-4)^3 b) f(x) = ((2x+3)/(5x-2))^3 c) f(x) = (6x-7)^2 * (6x^2+2)^3 d) f(x) = sin(2x^3 + 2x) e) f(x) = (x^3 + 2x + 1)/(2x+1)^2 f) f(x) = (2x)cos^2(3x) g) f(x) = cos(tan(x)) h) If y = x^2 sin 2x, then dy/dx =

Step-by-step solution with explanation

Final Answer

Step-by-step solution

Differentiate part a using Chain Rule

f(x) = (2x-4)^{-3}f'(x) = -3(2x-4)^{-4} \cdot \frac{d}{dx}(2x-4) = -3(2x-4)^{-4} \cdot 2

Rewrite the fraction as a negative exponent. Apply the chain rule: bring down the exponent, reduce it by 1, then multiply by the derivative of the inside. The derivative of (2x-4) is 2.

Write part a final derivative form

We keep it unsimplified as instructed. The -3 times 2 gives -6 out front, and the base stays as (2x-4) raised to the 4th power in the denominator.

Differentiate part b using Chain Rule

The outer function is something cubed. Chain rule says: bring down the 3, keep the inside, raise to the 2nd power, then multiply by the derivative of the inside quotient.

Apply Quotient Rule to the inside of part b

Quotient rule: (low times derivative of high, minus high times derivative of low) all over low squared. Derivative of (2x+3) is 2, derivative of (5x-2) is 5.

Combine for full part b derivative

Plug the quotient rule result back in. Do not simplify, as instructed. This is the complete derivative for part b.

Differentiate part c using Product Rule

Product rule: derivative of the first times the second, plus the first times the derivative of the second. Each derivative piece also needs the chain rule.

Apply Chain Rule to each factor in part c

Chain rule on the first factor: 2(6x-7) times derivative of inside (which is 6). Chain rule on the second factor: 3(6x^2+2)^2 times derivative of inside (which is 12x). Left unsimplified.

Differentiate parts d, e, f, g, h

d) Chain rule on sin: cos of the inside, times derivative of inside (6x^2+2). e) Quotient rule: (derivative of top)(bottom) minus (top)(derivative of bottom), over bottom squared; chain rule gives 2(2x+1)·2 for the bottom derivative. f) Product rule on (2x) and cos^2(3x), then chain rule twice on cos^2(3x): 2cos(3x)·(-sin(3x))·3. g) Chain rule: -sin(tan(x)) times derivative of tan(x) which is sec^2(x). h) Product rule: derivative of x^2 is 2x times sin(2x), plus x^2 times derivative of sin(2x) which is cos(2x)·2.

Understanding this problem

Learning Insight

Every problem here uses the Chain Rule, sometimes layered with Product or Quotient Rule. The Chain Rule says: when a function is INSIDE another function, you must multiply by the derivative of the inside. This is why each step has that extra multiplied factor — it accounts for how fast the inside is changing.

Quick Tip

For chain rule, ask yourself: 'What is the outermost operation?' Differentiate that first, keep the inside exactly the same, then multiply by the derivative of the inside. Work from outside to inside every time.

Common Mistake

Students forget to apply the chain rule to the inside function — for example, they write the derivative of sin(2x^3+2x) as just cos(2x^3+2x) and forget to multiply by the inside derivative (6x^2+2). Always multiply by the inside derivative!