Calculus

(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 = sin 2x, then dy/dx =

Step-by-step solution with explanation

Final Answer

a) b) c) d) e) f) g) h)

Step-by-step solution

1

Differentiate part a using Chain Rule

f(x) = (2x-4)^{-3}f'(x) = -3(2x-4)^{-4} \cdot (2)$
Rewrite the fraction as a negative exponent. Then apply the Chain Rule: bring down the power, reduce it by 1, and multiply by the derivative of the inside function , which is .
2

Differentiate part b using Chain Rule

f'(x) = 3\left(\frac{2x+3}{5x-2}\right)^{2} \cdot \frac{d}{dx}\left(\frac{2x+3}{5x-2}\right)\frac{d}{dx}\left(\frac{2x+3}{5x-2}\right) = \frac{(2)(5x-2)-(2x+3)(5)}{(5x-2)^{2}}$
Apply the Chain Rule to the outer cube, then use the Quotient Rule on the inside fraction. Do not simplify — leave it in this form.
3

Differentiate part c using Product and Chain Rules

f'(x) = (6x-7)^{2} \cdot 3(6x^{2}+2)^{2} \cdot (12x) + (6x^{2}+2)^{3} \cdot 2(6x-7) \cdot (6)$
Use the Product Rule: derivative of first times second plus first times derivative of second. Each factor also needs the Chain Rule applied to it.
4

Differentiate part d using Chain Rule

f'(x) = \cos(2x^{3}+2x) \cdot (6x^{2}+2)$
The outer function is sine, whose derivative is cosine. Multiply by the derivative of the inside, , which is .
5

Differentiate part e using Quotient Rule

f'(x) = \frac{(x^{3}+2x+1) \cdot 2(2x+1) \cdot (2) - (2x+1)^{2} \cdot (3x^{2}+2)}{\left[(2x+1)^{2}\right]^{2}}$
Apply the Quotient Rule: (denominator × derivative of numerator minus numerator × derivative of denominator) all over denominator squared. The denominator needs the Chain Rule for its derivative.
6

Differentiate part f using Product and Chain Rules

f'(x) = (2x) \cdot 2\cos(3x) \cdot (-\sin(3x)) \cdot (3) + \cos^{2}(3x) \cdot (2)$
Use the Product Rule on and . For , apply the Chain Rule twice: power rule gives , then derivative of gives .
7

Differentiate part g using Chain Rule twice

f'(x) = -\sin(\tan(x)) \cdot \sec^{2}(x)$
The outer function is cosine, whose derivative is negative sine. Multiply by the derivative of the inside , which is .
8

Differentiate part h using Chain Rule

\frac{dy}{dx} = \cos(2x) \cdot (2)$
The outer function is sine, whose derivative is cosine. Multiply by the derivative of the inside , which is .

Understanding this problem

Learning Insight

Every one of these problems uses the Chain Rule — the big idea is that when a function is 'nested' inside another, you work from the outside in, multiplying derivatives at each layer. The Product and Quotient Rules handle two functions multiplied or divided, but the Chain Rule handles functions composed inside each other. Most real-world functions require combining all three rules together.

Quick Tip

Remember the Chain Rule as 'outside-inside': take the derivative of the outer shell first (leaving the inside untouched), then multiply by the derivative of the inside. Write it out in two separate pieces so you don't lose track.

Common Mistake

Students forget to apply the Chain Rule to the inside function — for example, they differentiate as just and forget to multiply by from the inner layers.