Math Help - derivatives...blah

1. derivatives...blah

Find the derivative of the function below.

how would i do this one?

2. Originally Posted by mathaction
Find the derivative of the function below.

how would i do this one?
Hello,

first rearrange the equation:

$g(w) = w^{3.3} \cdot 7^{-w}$

You are supposed to know that $7^{-w} = e^{-\ln(7) \cdot w}$

$g(w) = w^{3.3} \cdot e^{-\ln(7) \cdot w}$. Now use product rule to derivate this function:

$g'(w) = e^{-\ln(7) \cdot w} \cdot 3.3 \cdot w^{2.3} + w^{3.3} \cdot e^{-\ln(7) \cdot w} \cdot (-\ln(7))$ . Factor this term: The common factor is $e^{-\ln(7) \cdot w} = 7^{-w}$

$g'(w) = 7^{-w} \cdot \left( 3.3 \cdot w^{2.3} + w^{3.3} \cdot (\ln(7) \right) = 7^{-w} \cdot w^{2.3} \left( 3.3 + w \cdot (\ln(7) \right)$

If f(x) = (7x + 8)(6x - 5), find f '(x) and f ''(x).

4. Originally Posted by mathaction

If f(x) = (7x + 8)(6x - 5), find f '(x) and f ''(x).
hi,

1. rearrange and use product rule:

$y(t) = \sqrt{t} \cdot (t^2+9)$

$y'(t)=(t^2+9) \cdot \frac12 \cdot t^{-\frac12} + \sqrt{t} \cdot 2t = \frac{(t^2+9) + 4t^2}{2\sqrt{t}} = \frac{5t^2+9}{2\sqrt{t}}$

2. Don't expand the brackets.

$f(x) = (7x+8)(6x-5)~\overrightarrow{\text{ derivate }}$ $f'(x) = 7(6x-5) + 6(7x+8)~\overrightarrow{\text{ derivate }} f''(x) = 7 \cdot 6+6 \cdot 7$

3. Do yourself and do us a favour: If you have a new question start a new thread because nobody will notice "from outside this thread" that you have an other problem to solve.

5. Hello, mathaction!

$y(t) \;=\;\frac{t^{\frac{1}{2}}}{t^2+9}$
Evidently, earboth despises the Quotient Rule . . .

We have: . $y'(t)\;=\;\frac{(t^2+9)\cdot\frac{1}{2}t^{\text{-}\frac{1}{2}} \,- \,t^{\frac{1}{2}}\cdot2t}{(t^2+9)^2} \;=\;\frac{\frac{1}{2}t^{\text{-}\frac{1}{2}}(t^2+9) \,- \,2t^{\frac{3}{2}}}{(t^2+9)^2}$

Multiply top and bottom by $2t^{\frac{1}{2}}\!:\;\;y'(t)\;=\;\frac{(t^2+9) - 4t^2}{2t^{\frac{1}{2}}(t^2+9)}$

Therefore: . $y'(t)\;=\;\frac{9-3t^2}{2\sqrt{t}\,(t^2+9)^2}$

If $f(x) \:= \:(7x + 8)(6x - 5)$, .find $f'(x)$ and $f''(x)$

With such a simple product, I would multiply it out . . .

$\begin{array}{ccc}f(x) & = & 42x^2 + 13x - 40 \\
f'(x) & = & 84x + 13 \\ f''(x) & = & 84 \end{array}$