dy/dx

[jmsic]

im used to always writting f prime as my form for derivatives but my teacher put dy/dx on a practice test and i was like.... uhhh... so im confused as to what to do...

so the problems are:

Given x[supflj25l1]2[/supflj25l1] + xy = 5 find d[supflj25l1]2[/supflj25l1]y/dx[supflj25l1]2[/supflj25l1] in terms of x and y.

and

Find dy/dx given that y[supflj25l1]3[/supflj25l1] + y[supflj25l1]2[/supflj25l1] - 5y - x[supflj25l1]2[/supflj25l1] = -4

and

At time t = 0, a diver jumps from a diving board that is 32 feet above the water. The position of the diver is given by s(t)= -16t[supflj25l1]2[/supflj25l1]+16t+32 where s is measured in feet and t is measured in seconds. Find the diver’s velocity when he lands in the water

i know people on here appreciate it more if you try the problems out first but i have NO idea how to do ANY of those three and i looked through my textbook to see if there were any odd numbered problems that looked similiar to these so i could try to figure it out but couldn't find any.

please help meeeeee T__T

 

[gnozahs]

Find dy/dx given that y[sup:2m7ggyvp]3[/sup:2m7ggyvp] + y[sup:2m7ggyvp]2[/sup:2m7ggyvp] - 5y - x[sup:2m7ggyvp]2[/sup:2m7ggyvp] = -4

I'm not that good in calculus but I would do this.

3y^2 dy/dx + 2y dy/dx - 5 dy/dx - 2x = 0

 

[soroban]

Hello, jmsic!

Your teacher is using alternate symbols for derivatives.

. . \(\displaystyle \begin{array}{c||c|c|c|c }\text{Function} & \text{1st der.} & \text{2nd der.} & \text{3rd der.} & \hdots\\ \hline\hline f(x) & f'(x) & f''(x) & f'''(x) & \hdots \\ \hline \\[-3mm] y & \frac{dy}{dx} & \frac{d^2y}{dx^2} & \frac{d^3y}{dx^3} & \hdots \\ \\[-3mm] \hline \end{array}\)

\(\displaystyle \text{Given: }\: x^2 + xy = 5,\;\;\text{ find }\frac{d^2y}{dx^2}\text{ in terms of }x\text{ and }y.\)

\(\displaystyle \text{We are given: }\:x^2 + xy \:=\:5\)


\(\displaystyle \text{Differentiate implicitly: }\:2x + x\!\cdot\!\frac{dy}{dx} + 1\!\cdot\! y \:=\:0\)


\(\displaystyle \text{Solve for }\frac{dy}{dx}\!:\quad\frac{dy}{dx} \;=\;\frac{-2x-y}{x} \quad\Rightarrow\quad \boxed{\frac{dy}{dx}\;=\;-2 - \frac{y}{x}}\) .**


\(\displaystyle \text{Differentiate again: }\quad \frac{d^2y}{dx^2} \;=\;0 - \left(\frac{x\cdot\frac{dy}{dx} - y\cdot1}{x^2}\right)\)

. . \(\displaystyle \text{and we have: }\;\frac{d^2y}{dx^2} \;=\;\frac{y - x\!\cdot\!\frac{dy}{dx}}{x^2}\)

\(\displaystyle \text{We }have\text{ found the second derivative, but it is is terms of }x,\,y,\,\text{ and }\tfrac{dy}{dx}\)
. . \(\displaystyle \text{and they said "in terms of }x\text{ and }y\text{ (only)"}\)

\(\displaystyle \text{So we substitute for }\frac{dy}{dx}\!:\) .**

. . \(\displaystyle \frac{d^2y}{dx^2} \;=\;\frac{y -x\left(-2-\frac{y}{x}\right)}{x^2} \;=\;\frac{y + 2x + y}{x^2}\)

\(\displaystyle \text{Therefore: }\;\frac{d^2y}{dx^2} \;=\;\frac{2x+2y}{x^2}\)

 

[jmsic]

for finding dy/dx given that y3 + y2 - 5y - x2 = -4
my work so far (im not even sure if im right):
3d[supwutss5w]2[/supwutss5w]/dx[supwutss5w]2[/supwutss5w] + 2dy/dx - 5 -2x = 0
dy/dx (3dy/dx + 2) = 2x + 5
then i think you divide everything by 3dy/dx + 2 but then dy/dx is on the denominator... so im lost

 

[BigGlenntheHeavy]

Last one:

\(\displaystyle s(t) \ = \ -16t^{2}+16t+32, \ distance \ function, \ hence, \ when \ he \ hits \ the \ water, \ his \ distance \ is \ zero.\)

\(\displaystyle Time \ it \ takes \ to \ hit \ the \ water: \ 0 \ = \ -16t^{2}+16t+32, \ t \ = \ 2 \ sec.\)

\(\displaystyle s\' \ (t) \ = \ v(t) \ = \ -32t \ +16, \ v(2) \ = \ -48ft/sec.\)

 

[jmsic]

lol i figured the last one out but i still dont get the second one....