# Calculus n-th derivatives

• Jan 6th 2007, 04:31 PM
asnxbbyx113
Calculus n-th derivatives
Find a formula of f^(n)(x), if f(x)=(4+x)^(-1).

Find the third degree polynomial Q such that Q(1)=-1, Q'(1)=1, Q''(1)=8, and Q'''(1)=18.

Suppose that f(x)=g(x)/x^3 and that g is twice differentiable. Find f'' in terms of g, g', and g''.

Thanks!
• Jan 6th 2007, 04:37 PM
ThePerfectHacker
Quote:

Originally Posted by asnxbbyx113

Suppose that f(x)=g(x)/x^3 and that g is twice differentiable. Find f'' in terms of g, g', and g''.

That,
$f'=\frac{g'x^3-3x^2g'}{x^6}$
$f'=\frac{g'}{x^3}-\frac{3g'}{x^4}$
$f''=\frac{g''x^3-3x^2g'}{x^6}-\frac{3g''x^4-12x^3g'}{x^8}$.
But I am not going to simplify that.
• Jan 6th 2007, 07:42 PM
Soroban
Hello, asnxbbyx113!

Quote:

Find a formula for $f^{(n)}(x)$, if $f(x) = (4+x)^{-1}$

Start cranking out derivatives and look for a pattern . . .

$\begin{array}{cccccc}f(x) & = & (4 + x)^{-1} \\
f'(x) & = & \text{-}1(4 + x)^{-2} \\
f''(x) & = & 1\!\cdot\!2(4 + x)^{-3} \\
f'''(x) & = & \text{-}1\!\cdot\!2\!\cdot\!3(4 + x)^{-4} \\
f^{(4)}(x) & = & 1\!\cdot\!2\!\cdot\!3\!\cdot\!4(4 + x)^{-5} \\
\vdots & & \vdots \end{array}$

Therefore: . $f^{(n)}(x)\;=\;(\text{-}1)^n\,n!\,(4 + x)^{\text{-}(n+1)}$

• Jan 6th 2007, 08:07 PM
Soroban
Hello again, asnxbbyx113!

Quote:

Find the third-degree polynomial $Q(x)$ such that:
. . $Q(1)=-1,\;\;Q'(1)=1,\;\; Q''(1)=8,\;\;Q'''(1)=18$

Let $q(x) \;=\;ax^3 + bx^2 + cx + d$

Then: . $\begin{array}{cccc}Q(x) & = & ax^3 + bx^2 + cx + d \\ Q'(x) & = & 3ax^2 + 2bx + c \\ Q''(x) & = & 6ax + 2b \\ Q'''(x) & = & 6a\end{array}$

Since $Q'''(1) = 18\!:\;\;6a \:=\:18\quad\Rightarrow\quad a \,=\,3$

Since $Q''(1) = 8\!:\;\;6(3)(1) + 2b\:=\:8\quad\Rightarrow\quad b \,=\,\text{-}5$

Since $Q'(1) = 1\!:\;\;3(3)(1^2) + 2(\text{-}5)(1) + c \:=\:1\quad\Rightarrow\quad c\,=\,2$

Since $Q(1) = \text{-}1\!:\;\;(3)(1^3) + (\text{-}5)(1^2) + (2)(1) + d \:=\:\text{-}1\quad\Rightarrow\quad d\,=\,\text{-}1$

Therefore: . $Q(x) \:=\:3x^3 - 5x^2 + 2x - 1$