Hi all ..
problem in Calculus ( derivaives ) ( slope )
Hi all
plese I want check my answer , I don't know how i solve this queation but i try
thanks
i.
The derivative of y = sin(x) is
y' = cos(x)
Then, using x = pi/4, what you you get?
iv.
Using the product rule,
$\displaystyle \dfrac{dR}{dM} = M^2\left(-\dfrac13\right) + \left(\dfrac{C}{2} - \dfrac{M}{3}\right)\cdot 2M$
And I don't get what you got.
v. I think that the 'jerk' means in some way the derivative... then yes, it's good.
How did you get that?
Unknown008 has already given you the correct answer. Maybe you should show us your working so we can identify where you went wrong. (Also remember to use brackets)
To make question (iv) easier, expand the brackets!
$\displaystyle M^2 (\frac{C}{2} - \frac{M}{3}) = \dfrac{C \cdot M^2}{2} - \dfrac{M^3}{3}$
Now that the brackets are expanded, there is no need to use the product rule and you should be able to differentiate it with no problem.
haha a jerk is a derivative? i've not heard that one before.
edit: apparently it is so...well wikipedia says it is.
http://en.wikipedia.org/wiki/Jerk_%28physics%29
"Jerk" is a physics term. It is the third derivative of the position function: first derivative is velocity, second derivative is acceleration, third derivative is jerk. But, r-soy, you have the wrong sign. If s= 5cos(t), then s'= -5sin(t), s''= -5cos(t), and s'''= 5sin(t).
Using the product rule on $\displaystyle R= M^2(\frac{C}{2}- \frac{M}{3})$, $\displaystyle R'= 2M(\frac{C}{2}- \frac{M}{3})+ M^2(-\frac{1}{3}= MC- \frac{2M^2}{3}-\frac{M^2}{3}= MC- M^2$. r-soy, you dropped the "3" in the denominator.