Determine the derivative?

Question

Determine the derivative of f(x) (6 with only x under the square root sign -2)(5 with only x under the square root sign +7)

Answer 1

do u mean 1) [6(x)^-2]*[5(x)^7] ? or 2) [6(x)^-2] , 3) [5(x)^7]
then the ans is 1) -150(x)^-6.....
2) -12(x)^-3
3) 35(x)^6

Answer 2

i got 30 times the square root of x plus 16 all over the square root of x.

Answer 3

The people that answerd your questions either assumed that f(x) was wrong or they just cant derive. My appoligies for their stupidity.

Assume f(x) = u(x)*v(x)

product rule states that f'(x) =(u)*(dv/dx)+(v)*(du/dx)

okay so lets say..
u(x) = 6*x^0.5 - 2
v(x) = 5*x^0.5 - 7

plug in u(x) and v(x) into the equation above derives:

f'(x) = 30+16x^-0.5

Answer 4

if by this you mean

f(x) = (6sqrt(x) - 2)(5sqrt(x) + 7)

(d/dx)(f(x)) = f'(x)*g(x) + f(x)g(x)'

f'(x) = (3(x^((1/2) - 1))(5sqrt(x) + 7) + (6sqrt(x) - 2)((5/2)x^((1/2) - 1))

f'(x) = ((3(5sqrt(x) + 7))/(sqrt(x)) + (((5/2)(6sqrt(x) - 2))/(sqrt(x))

f'(x) = (15sqrt(x) + 21)/(sqrt(x)) + (15sqrt(x) - 5)/(sqrt(x))

f'(x) = (15sqrt(x) + 21 + 15sqrt(x) - 5)/(sqrt(x))

f'(x) = (30sqrt(x) + 16)/(sqrt(x))

f'(x) = ((sqrt(x))(30sqrt(x) + 16))/x

f'(x) = (30x + 16sqrt(x))/x

Answer 5

f(x)=(6 sqrt(x)-2)(5 sqrt(x) + 7)

Let
f(x)=g(x)h(x)

where
g(x)=(6 sqrt(x)-2)
g'(x)= 3/sqrt(x)

h(x)=5 sqrt(x) + 7
h'(x)=2.5/sqrt(x)

f'(x)=g'(x)h(x)+g(x)h'(x)
=>
f'(x)=
3*(5 sqrt(x)+7)/sqrt(x)+2.5*(6 sqrt(x)-2)/sqrt(x)=
{15sqrt(x)+21+15sqrt(x)-5}/sqrt(x)=
(30sqrt(x)+16)/sqrt(x)
=====
NOTE
deriv of sqrt(x)= (1/2)/sqrt(x)
Note sqrt(x)=x^.5
so derivative
is .5 x^(0.5-1)=.5x^(-0.5)
=.5/sqrt(x)
keyboard is hard to show, it's easier with Math symbols.

Answer 6

if you mean d/dx(6sqrt(x^-2))(5sqrt(x)+7)
then the dirivative is

(-3abs(1/x)(5sqrt(x) + 14))/x