Question

Find the derivative of y= ln(cos(ln θ))

A. -tan(ln θ)/θ
B. -tan(ln θ)
C. tan(ln θ)
D. tan(ln θ)/θ

Answer 1

A

Explanation:

the chain rule for derivatives:

f(g(x))' = f'(g(x))⋅g'(x)

in our case we have f(g(h(x))), so that

f(g(h(x)))' = f'(g(h(x))) × g(h(x))' =

= f'(g(h(x))) × g'(h(x)) × h'(x)

ln(x)' = 1/x

cos(x)' = - sin(x)

tan(x) = sin(x)/cos(x)

ln(cos(ln theta)' = (1/(cos(ln theta))) × -sin(ln theta) × 1/theta

= -sin(ln theta)/cos(ln theta) × 1/theta =

= -tan(ln theta)/theta

Answer 2

`A. -(\tan(\ln\theta))/(\theta)`

Explanation:

To find the derivative of y= ln(cos(ln θ)), we can use the chain rule twice. The chain rule states that the derivative of a composite function f(g(x)) is
`\sf f'\Bigl(g(x)\Bigr)\cdot g'(x)`.

Let u(x)=ln(x),

v(x)=cos(x), and

w(x)=ln(x)

Then,

`\sf y=u\biggl(v\Bigl(w(x)\Bigr)\biggr)`

To find y', we will need to find $u'$, $v'$, and $w'$.

`\sf u'(x)=(1)/(x)`

`\sf v'(x)=-sin(x)`

`\sf w'(x)=(1)/(\theta)`

Now, we can apply the chain rule twice:

`\begin{aligned} y' &= (d)/(dx)\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &= u'\biggl(v\Bigl(w(x)\Bigr)\biggr) \cdot (d)/(dx)\left[v\Bigl(w(x)\Bigr)\right] \\\\ &= u'\biggl(v\Bigl(w(x)\Bigr)\biggr) \cdot v'\Bigl(w(x)\Bigr) \cdot w'(x) \\\\ &= (1)/(\cos(\ln\theta)) \cdot (-\sin(\ln\theta)) \cdot (1)/(\theta) \\\\ &= (-\sin(\ln\theta))/(\theta\cos(\ln\theta)) \\\\ &= \boxed{-(\tan(\ln\theta))/(\theta)} \end{aligned}`

So,

the answer is:

`A. -(\tan(\ln\theta))/(\theta)`