How do you find the derivative of y=t^sqrt(t) ?

Question

step by step process is appreciated

Answer 1

y = t^(sqrt(t))

This is a function to a function, and can only be solved using logarithmic differentiation. This means taking the natural log (or ln) of both sides, and the differentiating implicitly.

ln(y) = ln(t^ (sqrt(t)) )

Note that ln(a^c) = c*ln(a), so we can move the sqrt(t) in front of the ln(t).

ln(y) = [sqrt(t)] ln(t)

Now, differentiate implicitly. The derivative of sqrt(t) is 1/[2sqrt(t)], so using the product rule on the right hand side,

(1/y)(dy/dt) = (1/[2sqrt(t)]) ln(t) + [sqrt(t)] (1/t)

Now, multiply both sides by y.

dy/dt = y (1/[2sqrt(t)]) ln(t) + [sqrt(t)] (1/t)

Since we know y = t^(sqrt(t))), we replace y with that.

dy/dt = [t^(sqrt(t)))] { (1/[2sqrt(t)]) ln(t) + [sqrt(t)] (1/t) }

I'll leave this up to you to simplify, as I've just gone through the major steps.

Answer 2

We will use the form
F(t) = e ^[G(t)]
Now if F(t) = t^sqrt(t)
Then F(t) = e^[ln(t) sqrt(t)]
G(t) = ln(t) sqrt(t) which is the exponent of e
the derivitive is
F'(t) = F(t) G'(t)
G'(t) = ln(t) / [2 sqrt(t)] + sqrt(t) / t
G'(t) = ln(t) / [2 sqrt(t)] + 2 /(2 sqrt(t))
G'(t) = [ ln(t) + 2] / [2 sqrt(t)]

so

F'(t) = t^sqrt(t) [(ln(t) + 2) /( 2 sqrt(t))]

Now to test it out. Lets use t=3 and delta = .0001

F'(3) = 5.99756..
F(3.0001) = 6.705591635.....
F(3) = 6.7049918538...

[F(t + delta) - F(t)] / delta = 5.99781...
This is good agreement for a delta of .0001

Answer 3

Write t^sqrt(t) as e^Ln(t^sqrt(t))= e^[sqrt(t)*Ln(t)] (it is y

So the derive is y*[sqrt(t)/t + Ln(t)/2sqrt(t)]