Question

How do you algebraically determine the derivative of y=(sqrt(x))(x+1)?

Answer 1

I got y' = (1/2)x^(-1/2), but the answer is y' = (3/2)x^(1/2) + (1/2)x^(-1/2)

Answer 2

\[y = \frac{ \sqrt{x} }{ x+1 }\]
well without using calculus you would need the rise over run which would be \[y' = \frac{ f(b) - f(a) }{b-a }\]

Answer 3

for b and a you can use any numbers in the domain

Answer 4

you may have see the formula like this before :
\[m=\frac{ y _{2} -y_{1}}{ x_{2}-x_{1} }\]

Answer 5

I think I got it, thanks guys.

Answer 6

What do you mean by 'algebraically determine?'

Answer 7

Meaning anything but graphically

Answer 8

If you want derivative, I'd recommend product rule.
This is differential calculus, yes?

Answer 9

Yes it is

Answer 10

For \[\large y=(\sqrt{x})(x+1) \]
is the function?

Answer 11

Product rule:
for y=u(x)*v(x)
y' = u(x)*v'(x) + v(x)*u'(x)