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avatar+25 

dont know what to do :(

 Oct 14, 2021
 #1
avatar+117821 
+1

Firstly you do not need to be given  y(5)=-26     You can work that out for yourself.

 

Like this, sub in x=5 and solve for y

\(5*5^2+5*5+5y=20\\ \text{divide through by 5}\\ 25+5+y=4\\ y=-26\)

 

So giving you that was probably a bit of a red herring.

 

\(5x^2+5x+xy=20\\ \text{differentiate both sides}\\ 10x+5+x\frac{dy}{dx}+1y=0\\ \text{Sub in x=5}\\ 50+5+5\frac{dy}{dx}+y=0\\ 5\frac{dy}{dx}=-55-y\\ \text{but we already know that when x=5, y=-26 so sub that in too}\\ 5\frac{dy}{dx}=-55+26\\ 5\frac{dy}{dx}=-29\\ \frac{dy}{dx}=-5.8\\ \)

 

y'(5) = -5.8

 

 

 

LaTex:

5x^2+5x+xy=20\\
\text{differentiate both sides}\\
10x+5+x\frac{dy}{dx}+1y=0\\
\text{Sub in x=5}\\
50+5+5\frac{dy}{dx}+y=0\\
5\frac{dy}{dx}=-55-y\\
\text{but we already know that when x=5, y=-26 so sub that in too}\\
5\frac{dy}{dx}=-55+26\\
5\frac{dy}{dx}=-29\\
\frac{dy}{dx}=-5.8\\

 Oct 14, 2021
 #2
avatar+117821 
+1

 

Try looking at this unit of work.  This is the first video of a set.

 

Khan acadamy is a great free resource. 

 

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-2/v/implicit-differentiation-1

 Oct 14, 2021

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