Question

match the functions and their derivatives: 1. ( y = sin(x) \tan(x) ) 2. ( y = cos(\tan(x)) ) 3. ( y = \tan(x) ) 4. ( y = cos^3(x) ) a. ( y = sin(x) + \tan(x) sec(x) ) b. ( y = -3 cos^2(x) sin(x) ) c. ( y = -sin(\tan(x)) / cos^2(x) ) d. ( y = 1 + \tan^2(x) ) submit answer next item

Explanation:

For \( y = \tan(x) \) (Function 3)

Step1: Recall derivative of \( \tan(x) \)

The derivative of \( \tan(x) \) is a standard derivative formula. We know that \( \tan(x)=\frac{\sin(x)}{\cos(x)} \), and using the quotient rule or the standard derivative formula, the derivative of \( \tan(x) \) is \( \sec^{2}(x) \). Also, from the trigonometric identity \( \sec^{2}(x)=1 + \tan^{2}(x) \).
So, if \( y=\tan(x) \), then \( y' = 1+\tan^{2}(x) \), which matches option D.

For \( y=\sin(x)\tan(x) \) (Function 1)

Step1: Use product rule

The product rule states that if \( y = u(x)v(x) \), then \( y'=u'(x)v(x)+u(x)v'(x) \). Let \( u = \sin(x) \) and \( v=\tan(x) \).

Step2: Find \( u' \) and \( v' \)

We know that \( u'=\cos(x) \) (derivative of \( \sin(x) \)) and \( v'=\sec^{2}(x) \) (derivative of \( \tan(x) \)). Also, \( \sec(x)=\frac{1}{\cos(x)} \), so \( \sec^{2}(x)=\tan(x)\sec(x) \) (since \( \tan(x)=\frac{\sin(x)}{\cos(x)} \) and \( \sec(x)=\frac{1}{\cos(x)} \), so \( \tan(x)\sec(x)=\frac{\sin(x)}{\cos^{2}(x)}=\sec^{2}(x)\sin(x) \)? Wait, no, let's re - calculate:
\( y=\sin(x)\tan(x)=\sin(x)\frac{\sin(x)}{\cos(x)}=\frac{\sin^{2}(x)}{\cos(x)} \)
Using quotient rule: \( y'=\frac{2\sin(x)\cos(x)\cos(x)-\sin^{2}(x)(-\sin(x))}{\cos^{2}(x)}=\frac{2\sin(x)\cos^{2}(x)+\sin^{3}(x)}{\cos^{2}(x)}=\sin(x)\frac{2\cos^{2}(x)+\sin^{2}(x)}{\cos^{2}(x)} \). Wait, maybe better to use product rule:
\( y'=\cos(x)\tan(x)+\sin(x)\sec^{2}(x) \)
Since \( \tan(x)=\frac{\sin(x)}{\cos(x)} \), \( \cos(x)\tan(x)=\sin(x) \), and \( \sec^{2}(x)=\tan(x)\sec(x) \) (because \( \sec(x)=\frac{1}{\cos(x)} \), so \( \sin(x)\sec^{2}(x)=\sin(x)\frac{1}{\cos^{2}(x)}=\tan(x)\frac{1}{\cos(x)}=\tan(x)\sec(x) \)). So \( y'=\sin(x)+\tan(x)\sec(x) \), which matches option A.

For \( y = \cos(\tan(x)) \) (Function 2)

Step1: Use chain rule

The chain rule states that if \( y = f(g(x)) \), then \( y'=f'(g(x))\cdot g'(x) \). Let \( u = \tan(x) \), so \( y=\cos(u) \).

Step2: Find \( f'(u) \) and \( g'(x) \)

The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \), and the derivative of \( u = \tan(x) \) with respect to \( x \) is \( \sec^{2}(x)=\frac{1}{\cos^{2}(x)} \).
So \( y'=-\sin(\tan(x))\cdot\sec^{2}(x)=-\frac{\sin(\tan(x))}{\cos^{2}(x)} \), which matches option C.

For \( y=\cos^{3}(x) \) (Function 4)

Step1: Use chain rule

Let \( u = \cos(x) \), so \( y = u^{3} \).

Step2: Find \( y' \) with respect to \( x \)

The derivative of \( y = u^{3} \) with respect to \( u \) is \( 3u^{2} \), and the derivative of \( u=\cos(x) \) with respect to \( x \) is \( -\sin(x) \).
By the chain rule, \( y'=3u^{2}\cdot(-\sin(x)) \). Substituting back \( u = \cos(x) \), we get \( y'=- 3\cos^{2}(x)\sin(x) \), which matches option B.

Answer:

1. \( y = \sin(x)\tan(x) \) matches A. \( y'=\sin(x)+\tan(x)\sec(x) \)
2. \( y=\cos(\tan(x)) \) matches C. \( y'=-\frac{\sin(\tan(x))}{\cos^{2}(x)} \)
3. \( y = \tan(x) \) matches D. \( y'=1 + \tan^{2}(x) \)
4. \( y=\cos^{3}(x) \) matches B. \( y'=-3\cos^{2}(x)\sin(x) \)