Find the derivative of the function.
r(t) = log_4(t + sqrt(t))
My answer (sorry, it's messy):
(1+(1/2sqrt(t)))(1/tln(4)+sqrt(t)ln(4))
recall that
y = loga u
y' = (1/lna)*(1/u)*(du/dy)
I will assume that you have
r(t) = log4 (t + √t)
r ' (t) = (1/ln4)*( 1/(t + √t) )*(1 + (1/2)t^(-1/2) )
= (1/ln4) (1 + 1/(2√t) )/(t + √t)
if we multiply top and bottom by 2√t
= (1/ln4) (2√t + 1)/(2t^(3/2) + 2t)
or variations of that
As I read your solution, it appears to be correct.
I think this is what you meant, the second term is the denomiator... (1+(1/2sqrt(t)))/( (tln(4)+sqrt(t)ln(4)) )
Yes, that's what I meant. :) Thank you both!
To find the derivative of the function r(t) = log_4(t + sqrt(t)), we can use the chain rule.
The chain rule states that if we have a composition of functions, f(g(t)), then the derivative of f(g(t)) with respect to t is given by the derivative of f with respect to g multiplied by the derivative of g with respect to t.
In this case, f(u) = log_4(u) and g(t) = t + sqrt(t).
To find the derivative of f(u) = log_4(u), we can use the natural logarithm identity:
d/dx(log_a(x)) = 1 / (x * ln(a))
Plugging in a = 4 and u = t + sqrt(t), we have:
d/du(log_4(u)) = 1 / (u * ln(4))
= 1 / ((t + sqrt(t)) * ln(4))
Next, we need to find the derivative of g(t) = t + sqrt(t). This can be done by applying the power rule and chain rule:
d/dt(t + sqrt(t)) = 1 + (1/2)sqrt(t)
Now, we can use the chain rule to find the derivative of r(t) = log_4(t + sqrt(t)):
r'(t) = (d/du(log_4(u))) * (d/dt(t + sqrt(t)))
= (1 / ((t + sqrt(t)) * ln(4))) * (1 + (1/2)sqrt(t))
Therefore, the derivative of the function r(t) = log_4(t + sqrt(t)) is:
r'(t) = (1 / ((t + sqrt(t)) * ln(4))) * (1 + (1/2)sqrt(t))
Note: It's a good practice to simplify the expression as much as possible, but I have provided the derivative in a more expanded form for clarity.