What are the derivaties of...?

Question

1. sqrt(2) / x^7
2. sqrt(x)*(x-1)
3. (sqrt(3)*u)+sqrt(3u)
4. (5root(t^2))+(2*sqrt(t^5))

In addition can someone please explain the general method of finding the derivaties for square roots? Thank you!

Answer 1

Note that √2, etc. are just constants, and don't need any special treatment just because they have a square root sign in them.
Where x appears underneath the square root, write it as power 1/2 and use the power rule and chain rule to get the derivative.
In the last question, I assume "5root" means the fifth root; write 5root(t^2) as (t^2)^(1/5) = t^(2/5).
1. d/dx (√(2) / x^7) = √2 d/dx(x^7)
= √2 . 7x^6.

2. d/dx √(x)*(x-1) = √(x) . d/dx(x-1) + d/dx (√(x)) . (x-1)
= x^(1/2).1 + 1/2 (x^-(1/2)).(x-1)
= x^-(1/2) . [x + (1/2) (x-1)]
= (3x - 2) / (2√x)

3. d/du (√(3).u + √(3u))
= d/du (√3) (u + u^(1/2))
= √3 (1 + 1/2u^(-1/2))
= √3 (1 + 1/(2√u))

4. d/dt (5root(t^2) + 2√(t^5))
= d/dt ((t^2)^(1/5) + 2.(t^5)^(1/2))
= d/dt (t^(2/5) + 2t^(5/2))
= (2/5)t^(-3/5) + 2.(5/2)t^(3/2)
= (2/5)t^(-3/5) + 5t^(3/2).

Answer 2

1) sqrt(2)/x^7

First, factor the constant.

sqrt(2) [ 1/x^7 ]

Express 1/x^7 as x^n.

sqrt(2) [ x^(-7) ]

Use the power rule.

sqrt(2) (-7)x^(-8)

2) sqrt(x) * (x - 1)

Either use the product rule or expand.

x^(1/2) (x - 1)
x^(3/2) - x^(1/2)
(3/2)x^(1/2) - (1/2)x^(-1/2)

Try the rest on your own.

Answer 3

Write it as if raised to an exponent of .5 or 1/2. Use the power rule when the unknown (x) is what is being raised.

(2^.5)/(x^7) = (2^.5)(x^-7). (2^.5) is just a prefactor; it remains unchanged in the derivative: (2^.5)(-7x^-8)