Can someone explain how the answer to this calculus problem?

Question

The object is to calculate the second and third derivative of this function:
y = 5/(8+x)

However, I'm very confused. I already know how to get multiple derivatives of functions such as f(x) = 2x^3+9x-7, and I know the power rule, quotient rule, and product rule. For this problem, however, I think I'll have to get the first derivative by using the quotient rule, but how do I get the second derivitive of that? Do I just implement use of the quotient rule twice? Please help me understand this!

Answer 1

Find the first derivative using the quotient rule (or any other way).



u = 5, du = 0

v = 8+x, dv = 1

v^2 = (8+x)^2



[(8+x)(0) - 5(1)]/(8+x)^2 = -5/(8+x)^2



Find the second derivative of the original equation by finding the derivative of the derivative



using the quotient rule, it would look like this:



u = -5, du = 0

v = (8+x)^2, dv = 2(8+x) = 16+2x

v^2 = (8+x)^4



[(8+x)^2(0) - (-5)(16+2x)]/ (8+x)^4

10(8+x) / (8+x)^4 = 10/(8+x)^3



You can find the 3rd derivative of the original function, by finding the derivative of the 2nd derivative.

Answer 2

Just implement the quotient rule twice.



Think of it this way. The first derivative of a function is just another function in its own right. To take the first derivative of that "new" function, you'd do the same thing you'd do to take the first derivative of any other function.



But the first derivative of a first derivative is the second derivative of the original function.



f'(x) = g(x)

g'(x) = (f'(x))' = f"(x)

Answer 3

Just keep doing the quotient rule:



f(x) = 5 / (8 + x)



f'(x) = ((8 + x) * 0 - 5 * 1) / (8 + x)^2 = -5 / (8 + x)^2



f''(x) = ((8 + x)^2 * 0 - -5 * 2 * (x + 8)) / (8 + x)^4

f''(x) = 10 / (8 + x)^3



f'''(x) = ((8 + x)^3 * 0 - 10 * 3 * (x + 8)^2) / (8 + x)^6

f'''(x) = (-30) / (x + 8)^4

Answer 4

dy/dx = dy/d(8+x) (why? Or you can use function of a function, and d/dx of 8+x is what?)



f' = -5/(8+x)^2



f'' = 10/(8+x)^3



OK?

Answer 5

acc to ur given table, water in tank after t seconds would be: water(t)=a hundred-(t^2) differentiate wrt t: fee of pass= -2t (- sign shows water is reducing) so answer of A: placed t=2 i.e 4 gallons according to 2nd answer of B: placed t=4 i.e 8 gallons according to sec

Answer 6

diff(5/(8+x), x) = - 5/(8+x)^2



diff(-5/(8+x)^2, x) = 10/(8+x)^3





diff(10/(8+x)^3, x) = - 30/(8+x)^4