Question

Differentiate the following function:
F(t) = (ln(t))^2 cos(t)

I thought it would be:
F'(t)=ln(t)[ln(t)sin(t)+(2cos(t)/t)]
but apparently, that's not the right answer? Can somebody tell me what it would be and how you would get it?

Answer 1

[ ln ( t )² ] '

Substitution:

u = ln ( t )

ln ( t )² = u²

Using the chain rule:

du / dt = du / du ∙ ( du / dt ) =

d ( u ² ) / du ∙ d [ ln ( t ) ] / dt =

2 u ∙ 1 / t = 2 ln ( t ) / t


[ cos ( t ) ] ' = - sin ( t )


Apply the Product Rule:

( f · g ) ′ = f ′ · g + f · g′​

where:

f = ln ( t )² , cos ( t ) , f ′ = 2 ln ( t ) / t , g′​ = - sin ( t )


( f ∙ g )' = 2 ln ( t ) / t ∙ cos ( t ) + ln ( t )² ∙ [ - sin ( t ) ] =

2 ln ( t ) / t ∙ cos ( t ) - ln ( t )² ∙ sin ( t ) =

ln ( t ) [ 2 cos ( t ) / t - ln ( t ) ∙ sin ( t ) ]

Answer 2

Use the chain rule and the product rule
y = (ln(t))^2 cos(t)
y' = (2 * lnt * 1/t) * cost - (lnt)^2 sint
= lnt (2/t cost - lnt sint)