Question

Calculus review help

Answer 1

The question

Answer 2

What I think I'm supposed to do is take the derivative of that equation and then plug in the x and y values from the point.
Is that right?

Answer 3

looks like implicit differentiation, the power rule and the product rule.

Answer 4

and then (yes) put in x=-1 and y=2 and solve for dy/dx

Answer 5

What about the =49 do I just ignore that or differentiate that too

Answer 6

Differentiate that as well :)
And make sure you don't go through the trouble of solving for dy/dx BEFORE plugging in the coordinate, it will be a huge pain.

Answer 7

\[6y ^{2}x+5=196x ^{3}\] is what I got after differentiating is that right?

Answer 8

the derivative of a constant 49 is 0

Answer 9

how did you get that ?

Answer 10

No. Differentiating just the y^5 portion will give you\[5y ^{4}dy\]

Answer 11

\[\frac{ dy }{ dx }(y ^{5}+3x ^{2}y ^{2}+5x ^{4}) = 20x ^{3}+6y ^{2}x\]

Answer 12

That is what I get for the first portion

Answer 13

So do I just plug in the x and y values into that ^

Answer 14

unfortunately you have to treat the y as a variable also

Answer 15

Oh I just did d/dx whoops

Answer 16

notice
\[ \frac{d}{dx} y^5 = 5 y^4 \frac{dy}{dx}\]
also
\[ \frac{d}{dx}x^5 = 5x^4 \frac{dx}{dx}= 5x^4\]

Answer 17

Okay so do i do that for all of the different terms?

Answer 18

yes. and notice the term with x and y requires the product rule:
d(u v) = u dv + v du

Answer 19

okay so you did the first one already

Answer 20

\[\frac{ d }{ dx }3x ^{2}y ^{2} = 6xy ^{2}\frac{ dx }{ dy }\]

Answer 21

you mean dx/dx (normally people don't bother to write it)
6 x y^2

but that is only one of the terms: you did 3y^2 d/dy x^2
you also have to do 3x^2 * d/dy y^2

Answer 22

3x^2*2y

Answer 23

yes, but remember the chain rule. you should also have dy/dx

Answer 24

in other words, you should get
\[ 6 x^2 y \frac{dy}{dx} \]

Answer 25

okay so now do i plug in the x and y or is there still more we have differentiate

Answer 26

what did you get after taking the derivative?

Answer 27

\[5y ^{4}\frac{ dy }{ dx }+6x ^{2}y \frac{ dy }{ dx }=-6xy ^{2}-20x ^{3}\]

Answer 28

ok

Answer 29

\[y(5y ^{4}+6x ^{2})\frac{ dy }{ dx }\]

Answer 30

I would not bother simplifying your original equation. rather, put in -1 for x and 2 for y

Answer 31

okay

Answer 32

So after i did it all this is what i got
\[\frac{ dy }{ dx }=\frac{ -2x(3y ^{2}+10x ^{2}) }{ y(5y ^{4}+6x ^{2}) }\]

Answer 33

Is that right for dy/dx?

Answer 34

In the denominator, you should have 5y^3 instead of 5y^4.

Answer 35

Everything else is correct.

Answer 36

Okay so now i just plug in the x and y values now?

Answer 37

if you put in the numbers first, the algebra is easier.

Answer 38

okay

Answer 39

I'm not getting one of the answer choices

Answer 40

Okay i made a simple error I'm good now Thank you everyone!!!

Answer 41

I got 11/23

Answer 42

\(\checkmark\)

Answer 43

You are correct.

Answer 44

this is a really sweet way to do this kinda thing

https://www.youtube.com/watch?v=xtgTckGMuWE

but maybe further down the line

Answer 45

Thanks Everyone!