Question

Implicit Differentiation
Differentiate the following please?
xy + 2x^2 - 4y = 4x^4 - 3x^3 + 4y - 5

Answer 1

\[y+xy'+4x-4y'=16x ^{3}-9x ^{2}+4y'\]

Answer 2

\[xy + 2x^2 - 4y = 4x^4 - 3x^3 + 4y - 5\]Differentiate both sides w.r.t. x
\[\frac{d}{dx}(xy + 2x^2 - 4y) = \frac{d}{dx}(4x^4 - 3x^3 + 4y - 5)\]\[y +x\frac{dy}{dx} +4x - 4\frac{dy}{dx} = 16x^3 - 9x^2 + 4\frac{dy}{dx} \]\[x\frac{dy}{dx} - 4\frac{dy}{dx} - 4\frac{dy}{dx} = 16x^3 - 9x^2-4x -y\]\[\frac{dy}{dx} (x- 8) = 16x^3 - 9x^2-4x -y\]\[\frac{dy}{dx}= \frac{16x^3 - 9x^2-4x -y}{ (x- 8) }\]

Answer 3

@Callisto can i know how do u differentiate xy?

Answer 4

xy= x'y +xy'= y +xy'

Answer 5

\[\frac{d}{dx}xy = y\frac{d}{dx}x + x\frac{d}{dx}y = y +x\frac{dy}{dx}\]

Answer 6

okay @Callisto let's say if i have -3xy^2, then how to differentiate?

Answer 7

\[\frac{d}{dx}(-3xy^2)=y^2\frac{d}{dx}(-3x)+ (-3x)\frac{d}{dx}y^2\]\[=y^2(-3)+ (-3x)(2y)\frac{dy}{dx}y=-3y^2 -6xy\frac{dy}{dx}\]