generalized power function rule.?

Question

differentiate (generalized power function rule)

y = 1/sqrt(3x^2 + 7)

the rule says:

f'(x) = n[g(x)]^n-1 g'(x)

i dont see how this applies to that example? could someone give me a hint, please lol

Answer 1

sqrt(x) = x^.5

1/x^a = x^-a

Using these you can rewrite the example as

y = (3x^2 + 7)^-.5

y' = -.5(3x^2 + 7)^-1.5 * (6x)

or

y' = -3x/(3x^2 + 7)^1.5

Answer 2

Well sqrt can be written as X^(1/2) so your equation would be 1/(3x^2+7)^(1/2) and finally if you moved the sqrt to the top it would be

1*(3x^2+7)^(-1/2)
so now n= -1/2

so f'(x)= -1/2 * 1(3x^2+7)^((-1/2) -1) * 6x=
-6x/(2 (3x^2+7)^(-3/2)
the 6x in the numerator comes from the chain rule, (3x^2+7)

Answer 3

1/sqrt(3x^2+7) = (3x^2+7)^(-0.5)
sqrt = power 0.5 and 1/sqrt = power -0.5
g'(x) = 6x
so f'(x) = (3x^2+7)^(-1.5) *6x

you can write (3x^2+7)^(-1.5) =1/sqrt(3x^2+7)^3

and the answer is f'(x) = 6x/sqrt(3x^2+7)^3