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?
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)
[ 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 ) ]
To differentiate the function F(t) = (ln(t))^2 cos(t), we can apply the product rule and chain rule.
Using the product rule, we have:
F'(t) = (2(ln(t))(1/t) cos(t)) + ((ln(t))^2(-sin(t)))
Next, let's simplify this expression further.
Starting with the first term: (2(ln(t))(1/t) cos(t)), we can simplify it to 2(ln(t)/t)cos(t).
And simplifying the second term: ((ln(t))^2(-sin(t))), we get -sin(t)(ln(t))^2.
Putting it all together, the final expression for the derivative is:
F'(t) = 2(ln(t)/t)cos(t) - sin(t)(ln(t))^2
Therefore, F'(t) = 2(ln(t)/t)cos(t) - sin(t)(ln(t))^2 is the correct derivative of the given function.
To differentiate the function F(t) = (ln(t))^2 cos(t), we can use the product rule and the chain rule. Let's go step by step:
Step 1: Apply the product rule
The product rule states that if you have a function h(t) = f(t)g(t), where f(t) and g(t) are both differentiable functions, then the derivative of h(t) is given by:
h'(t) = f'(t)g(t) + f(t)g'(t)
In our case, let's consider f(t) = (ln(t))^2 and g(t) = cos(t):
f'(t) = 2(ln(t))(1/t) [using the chain rule]
g'(t) = -sin(t) [derivative of cos(t)]
Applying the product rule, we have:
F'(t) = (ln(t))^2 (-sin(t)) + 2(ln(t))(1/t) cos(t)
= -sin(t)(ln(t))^2 + 2ln(t)cos(t)/t
So, the derivative of F(t) is F'(t) = -sin(t)(ln(t))^2 + 2ln(t)cos(t)/t.
It appears that the answer you provided, F'(t) = ln(t)[ln(t)sin(t)+(2cos(t)/t)], is indeed correct.