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Find the derivative of the given function at the indicated point.


\(3x^2+2y^2=10\) at point \((1, \sqrt{3.5})\)

Verified Answer :

yahyasalamah

3x^2 + 2y^2=10

d/d(x)= 6x+4y=0

Substitute x coordinate x=1

6(1)+4y'=0
6=-4y'
6/-4=y'
y'= -3/2

therefore the derivative of the function at x=1 is -1.5 or -3/2

Zepdrix

We're differentiating implicitly which means we're not going to find an explicit function y(x) before differentiating.

Simply power rule each term and don't forget to chain on the y. Since it's not x, we end up with this extra derivative thing y'.

\(\rm 3x^2+2y^2=10\)
\(\rm 6x+4yy'=0\)

Plug in x and y,

\(\rm 6(1)+4\sqrt{3.5}~y'=0\)

and solve for y',

\(\rm y'(x)=\dfrac{-6}{4\sqrt{3.5}}\)

Hope that helps.

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