Answer:

Step-by-step explanation:
use y = mx + b
slope(m) =
y-intercept(b) = 3
so,

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
The correct answer is: " √x − <span>2√b " .
</span>_________________________________________________________
The "conjugate" of " √x + 2√b " is: " √x − 2√b " .
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Explanation:
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In an expression with 2 (TWO) terms; that is, in a "binomial expression",
the "conjugate" of that expression refers to that very expression — with the "sign" in between those two terms—"reverse" (e.g. "minus" becomes "plus" ; or, "plus" becomes "minus" .) .
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→ So: We are given: " <span>√x + 2√b " .
</span>
→ Note that this is a "binomial expression" ;
→ that is, there are 2 (TWO) terms: " <span>√x " ; and: " 2√b " .
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To find the "conjugate" of the given binomial expression:
</span>→ " <span>√x + 2√b " ;
</span>→ We simply change the "+" {plus sign} to a "<span>−" {minus sign} ; and rewrite:
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</span>→ " √x − 2√b " ;
→ which is the "conjugate" ; and is the correct answer:
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→ " √x − 2√b " ; is the "conjugate" of the expression: " <span>√x + 2√b " .
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</span>→ {that is: " √x − 2√b " ; is the conjugate.}.
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