# Thread: Derivative of a Constant help

1. ## Derivative of a Constant help

Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.

Find h'(3) for:

(a) h(x) = 2f(x) - 5g(x)

(b) h(x) = f(x)g(x)

(c) h(x) = f(x)/g(x)

(d) h(x) = g(x)/(2 + f(x))

As far as I know, the derivative of a constant is always 0. So when I plug in the values of f(x) and g(x) to find h(3) I always get a constant, and so the derivative of that must be zero, right?

2. Hello, topher0805!

. . $h'(3)$ means: .find $h'(x)$, then substitute $x = 3$

Given that: . $f(3) = \text{-}5,\;\;f'(3) = \text{-}2,\;\;g(3) = \text{-}2,\;\;g'(3) = 3$

Find $h'(3)$ for:

$(a)\;\;h(x) \:= \:2\!\cdot\!f(x) - 5\!\cdot\!g(x)$

We have: . $h'(x) \;=\;2\!\cdot\!f'(x) - 5\!\cdot\!g(x)$

$\text{Then: }\;h'(3) \;=\;2\!\cdot\!\underbrace{f'(3)}_{\downarrow} - 5\!\cdot\!\underbrace{g'(3)}_{\downarrow}$
- . - . . $h'(3) \;=\;\;\;2(\text{-}2) \;- \;5(3) \;\;=\;\;-4-15 \;\;=\;\;-19$

3. Originally Posted by topher0805
Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.

Find h'(3) for:

[snip]

(d) h(x) = g(x)/(2 + f(x))
Using the quotient rule:

$h'(x) = \frac{g'(x) (2 + f(x)) - f'(x) g(x)}{(2 + f(x))^2}$

$\Rightarrow h'(3) = \frac{g'(3) (2 + f(3)) - f'(3) g(3)}{(2 + f(3))^2}$

$\Rightarrow h'(3) = \frac{(3) (2 - 5) - (-2) (-2)}{(2 - 5)^2} = .....$

4. Originally Posted by topher0805
Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.

Find h'(3) for:

(a) h(x) = 2f(x) - 5g(x)
f(x) and g(x) are not constants, they are functions. So:

Initial equation of problem a
$h(x)=2f(x)-5g(x)$

then
$\frac d{dx}h(x)=\frac d{dx}[2f(x)-5g(x)]$

simplify
$\frac d{dx}h(x)=\frac d{dx}[2f(x)]-\frac d{dx}[5g(x)]$

Use Newtonian Notation
$h\prime (x)=2f\prime (x)-5g\prime (x)$

then
$h\prime (3)=2f\prime (3)-5g\prime (3)$

and substitute values
$h\prime (3)=2(-2)-5(3)$

simplify
$h\prime (3)=-4-15 = -19$

Can you do the rest?

edit: dang, I got doubly owned >.<

5. Originally Posted by Soroban
Hello, topher0805!

. . $h'(3)$ means: .find $h'(x)$, then substitute $x = 3$

We have: . $h'(x) \;=\;2\!\cdot\!f'(x) - 5\!\cdot\!g(x)$

$\text{Then: }\;h'(3) \;=\;2\!\cdot\!\underbrace{f'(3)}_{\downarrow} - 5\!\cdot\!\underbrace{g'(3)}_{\downarrow}$
- . - . . $h'(3) \;=\;\;\;2(\text{-}2) \;- \;5(3) \;\;=\;\;-4-15 \;\;=\;\;-19$

How did you line that up so well without arrays?

6. You can also use the aligned environment:

\begin{aligned}
h'(3) &= 2 \cdot \underbrace {f'(3)}_ \downarrow - \,5 \cdot \underbrace {g'(3)}_ \downarrow\\
&= ~~2( - 2) ~~- ~\,5(3)\\
&= - 4 - 15\\
&= - 19.
\end{aligned}

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Soroban uses the following: color beige, it's the perfect stealth to do what Soroban did.

7. Makes sense, of course, if MHF ever gets more skins, it won't be nearly so effective.