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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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P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{z^2}{2}}\\\\\text{where,}\\\\z = \frac{x-\mu}{\sigma}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-100)^2}{450}}

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z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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