DERivatives?

Question

Determine the derivative of f(x)=(6 square root sign with x under it minus 2)(5 square root sign with x under it +7)

Answer 1

I'm not sure what you have
1. d(u+v)=du + dv
or
2. d(uv)=vdu+udv (product rule)

if f(x)=[6(x)^1/2 -2][5(x)^1/2 +7]

either way using d((u)^1/2)=1/2[(u)^(-1/2)]*du or
d((u)^1/2)=1/2 / [(u)^(1/2)]*du


if f(x)=[6(x)^1/2 -2][5(x)^1/2 +7]
then f'(x)=][5(x)^1/2 +7][3/(x)^1/2]+[6(x)^1/2 -2][5/(2(x)^1/2)]

I hope this is what you asked for.

Answer 2

if I got your question right, square root being to the exponent one half. "^0.5"
then
f(x)=6 (X-2)^0.5 * 5 (x+7)^0.5
= 30 (X-2)^0.5 * (x+7)^0.5
so derivative is
f'(x)= 15(X-2)^-0.5 * (x+7)^0.5 + 15(X-2)^0.5 * (x+7)^-0.5
your final answer after simplifying should be something like: 15(2x + 5) all over the square root of (x-2)(x+7)

Answer 3

I understand f(x) = 6V(x-2) times 5V(x+7)
It is f(x) = 30V(x^2+5x-14) = 30 times (x^2+5x-14)^(1/2)
f'(x) = 30 times (x^2+5x-14)^(1/2 - 1) times (2x +5)
So f'(x) = 30(2x +5)/V(x^2+5x-14)