[ 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 ) ]
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?
2 answers
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)
y = (ln(t))^2 cos(t)
y' = (2 * lnt * 1/t) * cost - (lnt)^2 sint
= lnt (2/t cost - lnt sint)