Question

differentiate.
$y = \sec(\theta)\tan(\theta)$
$y=$

1. - / 1 points

find the derivative of the function.
$f(t)=t\cos(t)+t^{2}\sin(t)$
$f(t)=$

Explanation:

Step1: Recall product - rule

The product - rule states that if \(y = u\cdot v\), then \(y'=u'v + uv'\). For \(y=\sec(\theta)\tan(\theta)\), let \(u = \sec(\theta)\) and \(v=\tan(\theta)\).

Step2: Find derivatives of \(u\) and \(v\)

The derivative of \(\sec(\theta)\) is \(\sec(\theta)\tan(\theta)\), and the derivative of \(\tan(\theta)\) is \(\sec^{2}(\theta)\).

Step3: Apply product - rule

\(y'=\sec(\theta)\tan(\theta)\cdot\tan(\theta)+\sec(\theta)\cdot\sec^{2}(\theta)\).
Simplify to get \(y'=\sec(\theta)\tan^{2}(\theta)+\sec^{3}(\theta)=\sec(\theta)(\tan^{2}(\theta)+\sec^{2}(\theta))\).

For \(f(t)=t\cos(t)+t^{2}\sin(t)\):

Step1: Apply sum - rule of differentiation

If \(y = g(t)+h(t)\), then \(y'=g'(t)+h'(t)\). Here \(g(t)=t\cos(t)\) and \(h(t)=t^{2}\sin(t)\).

Step2: Apply product - rule to \(g(t)\)

Let \(u = t\) and \(v=\cos(t)\). Then \(u' = 1\) and \(v'=-\sin(t)\). By product - rule \(g'(t)=1\cdot\cos(t)+t\cdot(-\sin(t))=\cos(t)-t\sin(t)\).

Step3: Apply product - rule to \(h(t)\)

Let \(u = t^{2}\) and \(v=\sin(t)\). Then \(u' = 2t\) and \(v'=\cos(t)\). By product - rule \(h'(t)=2t\sin(t)+t^{2}\cos(t)\).

Step4: Find \(f'(t)\)

\(f'(t)=g'(t)+h'(t)=\cos(t)-t\sin(t)+2t\sin(t)+t^{2}\cos(t)=\cos(t)+t\sin(t)+t^{2}\cos(t)\).

Answer:

For \(y = \sec(\theta)\tan(\theta)\), \(y'=\sec(\theta)(\tan^{2}(\theta)+\sec^{2}(\theta))\)
For \(f(t)=t\cos(t)+t^{2}\sin(t)\), \(f'(t)=\cos(t)+t\sin(t)+t^{2}\cos(t)\)