What is the derivative of y=(sqrt(x))(x-1)??

Answer 1

Hi dear jc ;

if
y = (sqrt(x))(x-1) = (√x) * ( x-1)

y' = [(x-1) / (2√x) ] + √x
I will tell you how;
if f(x) = u(x) * v(x)
f'(x) = u'(x) * v (x) + u(x) *v'(x)

if u(x) = √x so ,u'(x) = 1/(2√x)

If v(x) = x-1 so ,v'(x) = 1

f'(x) = u'(x) * v (x) + u(x) *v'(x)
f'(x) = [1/(2√x) ]* x-1 ] + [1 * √x ] = [(x-1) / (2√x) ] + √x

It is your correct answer.
Good Luck dear.

Answer 2

Rewrite this as y = x^(1/2) * (x - 1)

This is the same as y = x^ (3/2) - x^(1/2)

The derivative is y' = (3/2)x^(1/2) - (1/2)x^(-1/2)

This is the same as y = 3(sqrt(x))/2 - 1/(2sqrt(x))

Answer 3

Use the product rule:

d/dx √x(x-1) = (x-1)/(2√x) + √x

Answer 4

y = (sqrt(x))(x - 1)
y = (x^(1/2))(x - 1)

y' = ((x^(1/2)')(x - 1)) + ((x^(1/2))(x - 1)')
y' = ((1/2)(x^((1/2) - 1))) + ((x^(1/2))(1))
y' = ((1/2)x^(-1/2)) + (x^(1/2))
y' = (1/(2x^(1/2))) + (x^(1/2))
y' = (1 + 2x)/(2x^(1/2))

y' = (1 + 2x)/(2x^(1/2))

Answer 5

y'=sqrt(x)+ (x-1)/2sqrt(x)