tough calculus problem please help?

Question

Find a formula for the nth derivative of the function 1/(4+9*x)

Answer 1

The derivative of u^k is k u^(k-1) du, so

f(x) = (4 + 9x)^-1

f'(x) = (-1)(4 + 9x)^-2 (9)

f''(x) = (-2)(-1)(4 + 9x)^-3 (9)(9)

...

n-th derivative is (-n)(- (n-1))...(-2)(-1)(4 + 9x)^-(n + 1) 9^n
or (-1)^n n! 9^n / ((4 + 9x)^(n+1))

Answer 2

It's all about finding a pattern, and then verifying it by induction.

f(x) = 1/(4 + 9x) = (4 + 9x)^(-1)

f'(x) = (-1)(4 + 9x)^(-2) (9), OR
f'(x) = (-1)(9) (4 + 9x)^(-2)

f''(x) = (-1)(-2)(9) (4 + 9x)^(-3) (9), OR
f''(x) = (-1)(-2) (9^2) (4 + 9x)^(-3)

f'''(x) = (-1)(-2)(-3) (9^3) (4 + 9x)^(-4)

Here's what the pattern appears to be:
Every odd derivative is negative (shown from the multiplication of (-1) and then (-2) and then (-3)). Therefore, the nth derivative is going to have (-1)^n.

The terms seem to accumulate: The 2nd derivative goes 1*2, and the third derivative goes 1*2*3. There's a function that acts this way; the factorial function, or n!. Therefore, the nth derivative is going to have n!.

We also have a power of 9, for the nth derivative; that is, 9^1 for the first derivative, 9^2 for the second, and so forth. 9^n would represent this.

Lastly, the power of (4 + 9x) is (-2) for the 1st derivative, (-3) for the 2nd derivative, so in general, it is -(n + 1). So the formula is also going to contain (4 + 9x)^(-(n + 1)).

Let's combine all of these together.

f(x) = 1/(4 + 9x). The formula for the n-th derivative is

f^(n)(x) = (-1)^n (n!) (9^n) (4 + 9x)^(-(n + 1)), OR, put as a fraction,

f^(n)(x) = (-1)^n (n!) (9^n) [1/(4 + 9x)^(n + 1)]

Answer 3

every time you take a derivative, you decrease the exponent of (4+9x) by 1, and multiply by both 9 and the previous exponent. The expression is

f'n(x) = (-9)^n * n! * (4+9x)^(-n-1)
or f'n (x) = (-9)^n * n! / [4+9x]^ (n+1)

If you're still not sure, try taking the first few derivatives,
1st: -1* 9 / (4+9x)^2
2nd: 2* 9^2 / (4+9x)^3
3rd: -6 * 9^3 / (4+9x)^4
4th: 24 *9^4 / (4+9x)^5
and so on.

Answer 4

Let u = 4+9x.
d^n f(x)/du^n
= (-1)^n n!/u^(n+1)

d^n f(x)/dx^n
= (d^n f(x) /du^n)(du/dx)^n
= (-9)^n n!/(4+9x)^(n+1)