Derivatives In Exercises \( 1-12 \), find \( d y / d x \) \( \begin{array}{ll}\text { 1. } y=-10 x+3 \cos x & \text { 2. } y=\frac{3}{x}+5 \sin x \\ \text { 3. } y=\csc x-4 \sqrt{x}+7 & \text { 4. } y=x^{2} \cot x-\frac{1}{x^{2}} \\ \text { 5. } y=(\sec x+\tan x)(\sec x-\tan x) \\ \text { 6. } y=(\sin x+\cos x) \sec x & \end{array} \)

We are given several functions y in terms of x and are asked to find dy/dx for each. 1. y = –10x + 3 cos x   • The derivative of –10x is –10.   • The derivative of 3 cos x is –3 sin x (since d/dx[cos x] = –sin x).   Thus, dy/dx = –10 – 3 sin x. 2. y = 3/x + 5 sin x   • Write 3/x as 3x⁻¹. Its derivative is –3x⁻² = –3/x².   • The derivative of 5 sin x is 5 cos x.   Thus, dy/dx = –3/x² + 5 cos x. 3. y = csc x – 4√x + 7   • The derivative of csc x is –csc x·cot x.   • Write √x as x^(1/2). Its derivative is (1/2)x^(–1/2), so the derivative of –4√x is –4(1/(2√x)) = –2/√x.   • The derivative of the constant 7 is 0.   Thus, dy/dx = –csc x·cot x – 2/√x. 4. y = x² cot x – (1/x²)   For the first term, use the product rule:    – Let u = x² (with u' = 2x) and v = cot x (with v' = –csc² x).    Then, d/dx[x² cot x] = 2x·cot x + x²(–csc² x) = 2x cot x – x² csc² x.   For the second term:    – Write –1/x² as –x^(–2); its derivative is –(–2)x^(–3) = 2x^(–3) = 2/x³.   Thus, dy/dx = 2x cot x – x² csc² x + 2/x³. 5. y = (sec x + tan x)(sec x – tan x)   Notice that (A + B)(A – B) = A² – B².   Let A = sec x and B = tan x, so y = sec² x – tan² x.   A fundamental trigonometric identity states that sec² x – tan² x = 1.   Thus, y = 1 and therefore dy/dx = 0. 6. y = (sin x + cos x) sec x   You can simplify:    – Expand: y = sin x·sec x + cos x·sec x.    – Since sec x = 1/cos x, cos x·sec x = 1, and sin x·sec x = sin x/cos x = tan x.   Thus, y = tan x + 1.   The derivative of tan x is sec² x, and the derivative of the constant 1 is 0.   Thus, dy/dx = sec² x. These are the derivatives for the given functions.