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pav-90 [236]
3 years ago
8

Implicit differentiation of 1/x +1/y=5 y(4)= 4/19 y'(4)=?

Mathematics
1 answer:
Aneli [31]3 years ago
5 0
\frac{1}{x} +\frac{1}{y} = 5\\\\x^{-1}+y^{-1}=5\\

Above, I changed the fraction form of x and y into exponential form so it is easier to see the differentiation. Now, we can differentiate:

-1x^{-2}+-1y^{-2}\frac{dy}{dx}=5\\\\\frac{-1}{x^2}-\frac{1}{y^2}\frac{dy}{dx}=5\\\\-\frac{1}{y^2}\frac{dy}{dx}=5+\frac{1}{x^2}\\\\\frac{dy}{dx}=-5y^2-\frac{y^2}{x^2}

Now that we have dy/dx, we can plug in the x, which is 4, and the y, which is 4/19. We know these values of x and y because your question stated y(4) = 4/19.

\frac{dy}{dx}=-5(\frac{4}{19})^2-\frac{(\frac{4}{19})^2}{(4)^2}\\\\\frac{dy}{dx}=-5(\frac{16}{361})-\frac{(\frac{16}{361})}{16}\\\\\frac{dy}{dx}=\frac{-80}{361}-\frac{1}{361}\\\\\frac{dy}{dx}=\frac{-81}{361}
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Tobeg says the equation 3x + 2y= 270 and y=4/5x + 20 intersect at the point (50,60) is this true how do you know
il63 [147K]

ANSWER

Yes it is very true


<u>EXPLANATION</u>

If the two equations intersect at (50,60) then this point must satisfy the two equations.


3x+2y=270---(1)

We substitute  (50,60)  in to erquation (1)


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y=\frac{4}{5}x+20 ---(2)


We now substitute  (50,60)  in to erquation (2) also


60=\frac{4}{5}(50)+20


60=40+20


60=60


Since the point satisfy all the two equations, it is true that they intersect at (60,50)




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