Find the derivative of the function below.
how would i do this one?
Hello,
first rearrange the equation:
$\displaystyle g(w) = w^{3.3} \cdot 7^{-w}$
You are supposed to know that $\displaystyle 7^{-w} = e^{-\ln(7) \cdot w}$
$\displaystyle g(w) = w^{3.3} \cdot e^{-\ln(7) \cdot w}$. Now use product rule to derivate this function:
$\displaystyle g'(w) = e^{-\ln(7) \cdot w} \cdot 3.3 \cdot w^{2.3} + w^{3.3} \cdot e^{-\ln(7) \cdot w} \cdot (-\ln(7))$ . Factor this term: The common factor is $\displaystyle e^{-\ln(7) \cdot w} = 7^{-w}$
$\displaystyle g'(w) = 7^{-w} \cdot \left( 3.3 \cdot w^{2.3} + w^{3.3} \cdot (\ln(7) \right) = 7^{-w} \cdot w^{2.3} \left( 3.3 + w \cdot (\ln(7) \right)$
hi,
1. rearrange and use product rule:
$\displaystyle y(t) = \sqrt{t} \cdot (t^2+9)$
$\displaystyle y'(t)=(t^2+9) \cdot \frac12 \cdot t^{-\frac12} + \sqrt{t} \cdot 2t = \frac{(t^2+9) + 4t^2}{2\sqrt{t}} = \frac{5t^2+9}{2\sqrt{t}}$
2. Don't expand the brackets.
$\displaystyle f(x) = (7x+8)(6x-5)~\overrightarrow{\text{ derivate }} $ $\displaystyle f'(x) = 7(6x-5) + 6(7x+8)~\overrightarrow{\text{ derivate }} f''(x) = 7 \cdot 6+6 \cdot 7$
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.
Hello, mathaction!
Evidently, earboth despises the Quotient Rule . . .$\displaystyle y(t) \;=\;\frac{t^{\frac{1}{2}}}{t^2+9}$
We have: .$\displaystyle y'(t)\;=\;\frac{(t^2+9)\cdot\frac{1}{2}t^{\text{-}\frac{1}{2}} \,- \,t^{\frac{1}{2}}\cdot2t}{(t^2+9)^2} \;=\;\frac{\frac{1}{2}t^{\text{-}\frac{1}{2}}(t^2+9) \,- \,2t^{\frac{3}{2}}}{(t^2+9)^2}$
Multiply top and bottom by $\displaystyle 2t^{\frac{1}{2}}\!:\;\;y'(t)\;=\;\frac{(t^2+9) - 4t^2}{2t^{\frac{1}{2}}(t^2+9)}$
Therefore: .$\displaystyle y'(t)\;=\;\frac{9-3t^2}{2\sqrt{t}\,(t^2+9)^2} $
If $\displaystyle f(x) \:= \:(7x + 8)(6x - 5)$, .find $\displaystyle f'(x)$ and $\displaystyle f''(x)$
With such a simple product, I would multiply it out . . .
$\displaystyle \begin{array}{ccc}f(x) & = & 42x^2 + 13x - 40 \\
f'(x) & = & 84x + 13 \\ f''(x) & = & 84 \end{array}$