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

Question

anonymous · Answer

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

jusaquikie · Answer

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

jusaquikie · Answer

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

jusaquikie · Answer

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

anonymous · Answer

I think I got it, thanks guys.

anonymous · Answer

What do you mean by 'algebraically determine?'

anonymous · Answer

Meaning anything but graphically

anonymous · Answer

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

anonymous · Answer

Yes it is

anonymous · Answer

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

anonymous · Answer

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