# Help Desperately Needed (chain rule)

• Feb 25th 2008, 06:27 PM
keyweez360
Help Desperately Needed (chain rule)
Working with the chain rule. Have two sets of problems, and I have no idea how to even get started. VERY confused/frustrated. Here are my two problem sets:

Code:

Suppose k '(1) = 4. (a) If f(x) = k(7x), find f '(1/7). f '(1/7)= (b) If g(x) = k(x+ 6), find g '(-5). g '(-5) = (c) If h(x) = k(x / 11), find h '(11). h '(11) =
and

Code:

Suppose f(2) = 4 and f '(2) = 7. Find the derivatives of the following functions. (a) g(x) = sqrt(f(x)) g '(2)= (b) h(x) = 1/f(x) h '(2)=
If anyone can help me out it would be greatly appreciated. I have to turn this in by 11pm.
• Feb 25th 2008, 07:26 PM
mr fantastic
Quote:

Originally Posted by keyweez360
Working with the chain rule. Have two sets of problems, and I have no idea how to even get started. VERY confused/frustrated. Here are my two problem sets:

Code:

Suppose k '(1) = 4. (a) If f(x) = k(7x), find f '(1/7). f '(1/7)= (b) If g(x) = k(x+ 6), find g '(-5). g '(-5) = (c) If h(x) = k(x / 11), find h '(11). h '(11) =
and

Code:

Suppose f(2) = 4 and f '(2) = 7. Find the derivatives of the following functions. (a) g(x) = sqrt(f(x)) g '(2)= (b) h(x) = 1/f(x) h '(2)=
If anyone can help me out it would be greatly appreciated. I have to turn this in by 11pm.

Chain rule: if $f(x) = g(h(x))$ then $f^{'}(x) = g^{'}(h(x)) \, h^{'}(x)$.

I'll do Q1 (a):

$f^{'}(x) = k^{'}(7x) \, (7x){'} = 7 k^{'}(7x)$.

Therefore $f^{'}(1/7) = 7 k^{'}(7 (1/7)) = 7 k^{'}(1) = (7) (4) = 28$.

As for Q2:

Again using the chain rule:

(a) $g^{'}(x) = \frac{1}{2} \frac{1}{\sqrt{f(x)}} \, f^{'}(x)$.

(b) $h^{'}(x) = - \frac{1}{[f(x)]^2} \, f^{'}(x)$.
• Feb 25th 2008, 07:30 PM
keyweez360
Fantastic!