Question

if = sqrtofx = sqrt of y =4,find y doble prime
a. sqrt of y over sqrt of x
b. sqrt of x over sqrt of y
c. negative sqrt of x over negative sqrt of y
d. negative sqrt of y over negative sqrt of x

Answer 1

\[find y'' if \sqrt x + \sqrt y =4 \]

Answer 2

We have
\[\sqrt x+\sqrt y=4\]
Let's differentiate this with respect to x
\[\frac{d}{dx}(\sqrt x+\sqrt y=4)\]
we get
\[ \frac{1}{2\sqrt x}+ \frac{1}{2\sqrt y} *y'=0\]
we have now
\[\frac{1}{2\sqrt y} *y'=-\frac{1}{2\sqrt x}\]
we have now
\[y'=-\frac{\sqrt y}{\sqrt x}\]
now let's differentiate this again with respect to x
\[y''=-(\frac{\sqrt x (\frac{d}{dx}\sqrt y)-\sqrt y (\frac{d}{dx} \sqrt x) }{x})\]
we get now
\[y''=-(\frac{\sqrt x (\frac{1}{2\sqrt y} *y')-\sqrt y (\frac{1}{2\sqrt x} )}{x})\]
we have \[y'=-\frac{\sqrt y}{\sqrt x}\]
so
\[y''=-(\frac{\sqrt x (\frac{1}{2\sqrt y} *-\frac{\sqrt y}{\sqrt x})-\sqrt y (\frac{1}{2\sqrt x} )}{x})\]
so we get finally
\[y''=-(\frac{-\frac{1}{2}-\sqrt y (\frac{1}{2\sqrt x} )}{x})\]
we get
\[y''=\frac{1+ \frac{\sqrt y}{\sqrt x} }{2x}\]

Answer 3

in y''' cancel 1 and 2x then the answer is