Differentiate the following with respect to x.

Here are the solutions for the differentiation problems:

1) $y = \sin 6x$ The student's answer is incorrect.

Step 1: Apply the chain rule. The derivative of $\sin(u)$ is $\cos(u) \cdot u'$. Here, $u = 6x$, so $u' = 6$. $\frac{dy}{dx} = \cos(6x) \cdot 6$ Step 2: Simplify the expression. $\frac{dy}{dx} = 6 \cos 6x$ The correct answer is $\boxed{6 \cos 6x}$.

2) $y = 3 \tan x$ The student's answer is correct.

Step 1: Use the constant multiple rule and the derivative of $\tan x$. The derivative of $\tan x$ is $\sec^2 x$. $\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\tan x)$ Step 2: Substitute the derivative. $\frac{dy}{dx} = 3 \sec^2 x$ The correct answer is $\boxed{3 \sec^2 x}$.

3) $y = \sin \frac{1}{2}x$

Step 1: Apply the chain rule. The derivative of $\sin(u)$ is $\cos(u) \cdot u'$. Here, $u = \frac{1}{2}x$, so $u' = \frac{1}{2}$. $\frac{dy}{dx} = \cos\left(\frac{1}{2}x\right) \cdot \frac{1}{2}$ Step 2: Simplify the expression. $\frac{dy}{dx} = \frac{1}{2} \cos\left(\frac{1}{2}x\right)$ The answer is \boxed{\frac{1{2} \cos\left(\frac{1}{2}x\right)}}.

4) $y = \cos(2x + 5)$

Step 1: Apply the chain rule. The derivative of $\cos(u)$ is $-\sin(u) \cdot u'$. Here, $u = 2x + 5$, so $u' = 2$. $\frac{dy}{dx} = -\sin(2x + 5) \cdot 2$ Step 2: Simplify the expression. $\frac{dy}{dx} = -2 \sin(2x + 5)$ The answer is $\boxed{-2 \sin(2x + 5)}$.

5) $y = \tan(3x - 1)$

Step 1: Apply the chain rule. The derivative of $\tan(u)$ is $\sec^2(u) \cdot u'$. Here, $u = 3x - 1$, so $u' = 3$. $\frac{dy}{dx} = \sec^2(3x - 1) \cdot 3$ Step 2: Simplify the expression. $\frac{dy}{dx} = 3 \sec^2(3x - 1)$ The answer is $\boxed{3 \sec^2(3x - 1)}$.

6) $y = 4 \tan 2x$

Step 1: Apply the constant multiple rule and the chain rule. The derivative of $\tan(u)$ is $\sec^2(u) \cdot u'$. Here, $u = 2x$, so $u' = 2$. $\frac{dy}{dx} = 4 \cdot \sec^2(2x) \cdot 2$ Step 2: Simplify the expression. $\frac{dy}{dx} = 8 \sec^2(2x)$ The answer is $\boxed{8 \sec^2(2x)}$.

7) $y = \sin x - \cos x$

Step 1: Differentiate each term separately. The derivative of $\sin x$ is $\cos x$. The derivative of $\cos x$ is $-\sin x$. $\frac{dy}{dx} = \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x)$ Step 2: Substitute the derivatives. $\frac{dy}{dx} = \cos x - (-\sin x)$ Step 3: Simplify the expression. $\frac{dy}{dx} = \cos x + \sin x$ The answer is $\boxed{\cos x + \sin x}$.

8) $y = 2 \cos x + 3 \tan x$

Step 1: Differentiate each term separately using the constant multiple rule. The derivative of $2 \cos x$ is $2(-\sin x) = -2 \sin x$. The derivative of $3 \tan x$ is $3(\sec^2 x) = 3 \sec^2 x$. $\frac{dy}{dx} = \frac{d}{dx}(2 \cos x) + \frac{d}{dx}(3 \tan x)$ Step 2: Substitute the derivatives. $\frac{dy}{dx} = -2 \sin x + 3 \sec^2 x$ The answer is $\boxed{-2 \sin x + 3 \sec^2 x}$.

9) $y = \cos 2x - \sin 2x$

Step 1: Differentiate each term using the chain rule. For $\cos 2x$: $u = 2x$, $u' = 2$. Derivative is $-\sin(2x) \cdot 2 = -2 \sin 2x$. For $\sin 2x$: $u = 2x$, $u' = 2$. Derivative is $\cos(2x) \cdot 2 = 2 \cos 2x$. $\frac{dy}{dx} = (-2 \sin 2x) - (2 \cos 2x)$ Step 2: Simplify the expression. $\frac{dy}{dx} = -2 \sin 2x - 2 \cos 2x$ $\frac{dy}{dx} = -2(\sin 2x + \cos 2x)$ The answer is $\boxed{-2(\sin 2x + \cos 2x)}$.

10) $y = \tan 2x + \sin 4x$

Step 1: Differentiate each term using the chain rule. For $\tan 2x$: $u = 2x$, $u' = 2$. Derivative is $\sec^2(2x) \cdot 2 = 2 \sec^2 2x$. For $\sin 4x$: $u = 4x$, $u' = 4$. Derivative is $\cos(4x) \cdot 4 = 4 \cos 4x$. $\frac{dy}{dx} = (2 \sec^2 2x) + (4 \cos 4x)$ The answer is $\boxed{2 \sec^2 2x + 4 \cos 4x}$.

11) $f(x) = 2x \sin \frac{x}{2}$

Step 1: Apply the product rule: $(uv)' = u'v + uv'$. Let $u = 2x$ and $v = \sin \frac{x}{2}$. Step 2: Find the derivatives of $u$ and $v$. $u' = \frac{d}{dx}(2x) = 2$. $v' = \frac{d}{dx}\left(\sin \frac{x}{2}\right) = \cos\left(\frac{x}{2}\right) \cdot \frac{1}{2} = \frac{1}{2} \cos\left(\frac{x}{2}\right)$. Step 3: Substitute into the product rule formula. $f'(x) = (2)\left(\sin \frac{x}{2}\right) + (2x)\left(\frac{1}{2} \cos \frac{x}{2}\right)$ Step 4: Simplify the expression. $f'(x) = 2 \sin \frac{x}{2} + x \cos \frac{x}{2}$ The answer is \boxed{2 \sin \frac{x{2} + x \cos \frac{x}{2}}}.

12) $f(x) = \frac{\sin 2x}{\cos 3x}$

Step 1: Apply the quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$. Let $u = \sin 2x$ and $v = \cos 3x$. Step 2: Find the derivatives of $u$ and $v$. $u' = \frac{d}{dx}(\sin 2x) = \cos(2x) \cdot 2 = 2 \cos 2x$. $v' = \frac{d}{dx}(\cos 3x) = -\sin(3x) \cdot 3 = -3 \sin 3x$. Step 3: Substitute into the quotient rule formula. $f'(x) = \frac{(2 \cos 2x)(\cos 3x) - (\sin 2x)(-3 \sin 3x)}{(\cos 3x)^2}$ Step 4: Simplify the expression. $f'(x) = \frac{2 \cos 2x \cos 3x + 3 \sin 2x \sin 3x}{\cos^2 3x}$ The answer is \boxed{\frac{2 \cos 2x \cos 3x + 3 \sin 2x \sin 3x{\cos^2 3x}}}.

13) $f(x) = \sin^3 x$

Step 1: Rewrite the function as $f(x) = (\sin x)^3$. Step 2: Apply the chain rule. Let $u = \sin x$, so $u' = \cos x$. The derivative of $u^3$ is $3u^2 \cdot u'$. $f'(x) = 3(\sin x)^2 \cdot \cos x$ Step 3: Simplify the expression. $f'(x) = 3 \sin^2 x \cos x$ The answer is $\boxed{3 \sin^2 x \cos x}$.

14) $y = \sqrt{\cos 2x}$

Step 1: Rewrite the function as $y = (\cos 2x)^{1/2}$. Step 2: Apply the chain rule. Let $u = \cos 2x$. Then $u' = -\sin(2x) \cdot 2 = -2 \sin 2x$. The derivative of $u^{1/2}$ is $\frac{1}{2}u^{-1/2} \cdot u'$. $\frac{dy}{dx} = \frac{1}{2}(\cos 2x)^{-1/2} \cdot (-2 \sin 2x)$ Step 3: Simplify the expression. $\frac{dy}{dx} = \frac{1}{2\sqrt{\cos 2x}} \cdot (-2 \sin 2x)$ $\frac{dy}{dx} = \frac{-\sin 2x}{\sqrt{\cos 2x}}$ The answer is \boxed{\frac{-\sin 2x{\sqrt{\cos 2x}}}}.

15) $y = \frac{1}{\sin x}$

Step 1: Rewrite the function as $y = (\sin x)^{-1}$. Step 2: Apply the chain rule. Let $u = \sin x$. Then $u' = \cos x$. The derivative of $u^{-1}$ is $-1u^{-2} \cdot u'$. $\frac{dy}{dx} = -1(\sin x)^{-2} \cdot \cos x$ Step 3: Simplify the expression. $\frac{dy}{dx} = -\frac{\cos x}{\sin^2 x}$ This can also be written as $-\cot x \csc x$. The answer is \boxed{-\frac{\cos x{\sin^2 x}}}.

16) $f(x) = \frac{1}{\cos x}$

Step 1: Rewrite the function as $f(x) = (\cos x)^{-1}$. Step 2: Apply the chain rule. Let $u = \cos x$. Then $u' = -\sin x$. The derivative of $u^{-1}$ is $-1u^{-2} \cdot u'$. $f'(x) = -1(\cos x)^{-2} \cdot (-\sin x)$ Step 3: Simplify the expression. $f'(x) = \frac{\sin x}{\cos^2 x}$ This can also be written as $\tan x \sec x$. The answer is \boxed{\frac{\sin x{\cos^2 x}}}.

Drop the next question! 📸