Step 1:
Start by putting

in front of each term
![\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5By%20cos%20x%5D%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B5x%5E2%5D%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%203y%5E2%5D)
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Step 2:
Deal with the terms in 'x' and the constant terms
![\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bycosx%5D%3D%2010x%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5B3y%5E2%5D%20%20)
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Step 3:
Use the chain rule for the terms in 'y'
![\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bycosx%5D%3D10x%2B6y%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%20)
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Step 4:
Use the product rule on the term in 'x' and 'y'


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Step 5:
Rearrange to make

the subject


![[cos(x) - 6y] \frac{dy}{dx}=10x + y sin(y)](https://tex.z-dn.net/?f=%5Bcos%28x%29%20-%206y%5D%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D10x%20%2B%20y%20sin%28y%29%20)

⇒ Final Answer
Answer: No unless the fbi cathches up to you mr.gieco
Step-by-step explanation:
Answer:
you can do a trial and improvement
Step-by-step explanation:
I believe none is the answer due to the fact it isn’t symmetrical. sorry if i am wrong. let me know if i am
Vertex is now at (-1,5)
for
y=a(x-h)^2+k
vertex is (h,k)
so veertex is (-1,5)
y=a(x-(-1))^2+5
y=a(x+1)^2+5
a is a constant, we will asssume that it is 1 because all the choices have 1
y=1(x+1)^2+5
y=(x+1)^2+5
2nd option