Derivatives?

Question

Can you explain how to do this problem
I have to find the Derivative of : squareroot(2t^2 + 5t)
Please esplain in detail if possible, thanks!

thanks for the quick answers and explanations!

Answer 1

Ok. The square root is like raising something to the 1/2 power. So we have:
(2t^2+5t)^(1/2)

Now, this satisfies the form x^n. The derivative of a function satisfying this form is:
n * (x)^(n - 1)

However, x itself is a function. Thus, we must use the Chain Rule. The chain rule, in this case, states that if f(x) = (g(x))^n, then f'(x) << n * (g(x))^(n - 1) * g'(x)

So, g(x) = 2t^2 + 5t.
Using what we know about x^n, the derivative of g(x) is:
g'(x) = 2(2(t)^1) + 5(1(t)^0) = 2(2(t)) + 5(1) = 4t + 5.

Plug that in for g'(x) and plug in g(x) = 2t^2 + 5t
and we get:

f'(t) = (1/2)(2t^2 + 5t)^(-1/2) * (4t + 5)

Derivative is the instantaneous rate of change of a function. Basically the slope of a function at any point x.

Answer 2

change sqrt to (2t^2 +5t)^1/2
then use the power rule and the chain rule:
1/2(2t^2 +5t)^-1/2 (2t^2+5t)'=
4t+5/(2(2t^2+5t)^1/2) factor out the t in the denominator:
4t+5/(2(t(2t+5))^1/2))=
4t+5/(2(sqrt t(2t+5)))

Answer 3

derivative (2t^2+5t)^0.5 =y since power ^1/2 = sqare root

derivative of u^n = n u^(n-1) du/dt

here n=1/2, n-1 = -1/2, du/dt = 4t+5

since a^-1/2 = 1/a^1/2

so result dy/dt = 1/2 (4t-5)/((2t^2+5t)^0.5)

Answer 4

f(t) = sqrt(2t^2 + 5t)

Note that the derivative of sqrt(x) is 1/[2sqrt(x)], so

f'(t) = 1/[2sqrt(2t^2 + 5t)] * (4t + 5)

{I used the chain rule.

f'(t) = (4t + 5) / [2sqrt(2t^2 + 5t)]