Questions: 7. If f(x)=sin x+tan x+(x^3/3), then f'(x)= A. -cos x-csc ^2 x+3 x^2 B. cos x+sec ^2 x+x^2 C. cos x-csc ^2 x-3 x^2 D. cos x+tan x sec x+x^2 8. If f(x)=cot x tan x, then f'(x)= A. cot x sec ^2 x-tan x csc ^2 x B. 1 C. -1 D. 0 9. If f(x)=(x^2-3 x) sec x, then f'(x)= A. sec x[(x^2-3 x) tan x+(2 x-3)] B. sec x[(x^2-3 x) tan x-(2 x-3)] C. (2 x-3) sec x tan x+(x^2-3 x) sec x D. (x^2-3 x) tan x+sec x(2 x-3)

Solution Steps

7. To find the derivative \( f'(x) \) of the function \( f(x) = \sin x + \tan x + \frac{x^3}{3} \), apply the derivative rules for each term separately: the derivative of \(\sin x\) is \(\cos x\), the derivative of \(\tan x\) is \(\sec^2 x\), and the derivative of \(\frac{x^3}{3}\) is \(x^2\).

8. For the function \( f(x) = \cot x \tan x \), simplify the expression first, as \(\cot x \tan x = 1\). The derivative of a constant is 0.

9. To find the derivative \( f'(x) \) of the function \( f(x) = (x^2 - 3x) \sec x \), use the product rule. The product rule states that the derivative of two functions \( u(x) \) and \( v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). Here, \( u(x) = x^2 - 3x \) and \( v(x) = \sec x \).

To find the derivative \( f'(x) \), we differentiate each term: \[ f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\tan x) + \frac{d}{dx}\left(\frac{x^3}{3}\right) \] Calculating each derivative gives: \[ f'(x) = \cos x + \sec^2 x + x^2 \] Thus, the expression simplifies to: \[ f'(x) = x^2 + \cos x + \sec^2 x \]

First, we simplify \( f(x) \): \[ f(x) = \cot x \tan x = 1 \] The derivative of a constant is: \[ f'(x) = 0 \]

Using the product rule: \[ f'(x) = \frac{d}{dx}(x^2 - 3x) \cdot \sec x + (x^2 - 3x) \cdot \frac{d}{dx}(\sec x) \] Calculating the derivatives: \[ \frac{d}{dx}(x^2 - 3x) = 2x - 3 \] \[ \frac{d}{dx}(\sec x) = \sec x \tan x \] Thus, we have: \[ f'(x) = (2x - 3) \sec x + (x^2 - 3x) \sec x \tan x \]

Final Answer