Question

Calculus: how do you find Find dy/dx by implicit differentiation? : 3xy-x^2-4=0

Answer 1

Well you have a product there....so we can start by using the product rule on

3xy

let f(x) = 3x and g(x) = y

\[f'(x)g(x) + f(x)g'(x)\]

Answer 2

And remember that since you are differentiating with respect to x...you cannot differentiate 'y' ...it would simply become y'

Answer 3

\[3xy-x ^{2}-4=0\]
diff.w.r.t.x
\[3\left( x \frac{ dy }{dx } +y*1\right)-2x=0,3x \frac{ dy }{dx }+3y=2x\]
\[\frac{ dy }{ dx }=\frac{ 2x-3y }{ 3x }\]

Answer 4

so we have

\[\frac{ d }{ dx } 3xy - x^2 - 4 = \frac{ d }{ dx }0\]
via the product rule
\[\frac{ d }{ dx } 3xy = 3\times y + 3x \times y'\]
via the power rule
\[\frac{ d }{ dx } -x^2 = -2x\]
via the constant rule
\[\frac{ d }{ dx } -4 = 0\]
and again via the constant rule
\[\frac{ d }{ dx } 0 = 0\]

So altogether we have

\[3y + 3xy' -2x = 0\]

Add 2x and subtract 3y from both sides because we want to isolate the y'

\[3xy' = -3y + 2x\]

And finally to solve for y' we divide both sides by 3x

\[y' = \frac{ -3y + 2x }{ 3x }\]

Answer 5

Good work @surjithayer ! medal for you!

Answer 6

yw