Differentiate with respect to x: $\displaystyle x^3siny - y^3cos = x^3$
I've never done this sort of differentiation before. Could someone please show me how it's done?
some good examples and explanation here ...
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html
go learn, try your problem, then receive feedback from your attempt.
Hello, xwrathbringerx!
You've never done Implicit Differentiation?
Then who assigned this problem to you?
Besides, you left out part of the problem . . .
Differentiate with respect to $\displaystyle x\!:\;\;x^3\sin y - y^3\cos {\color{red}?} \:=\: x^3$
I'll assume the problem is: .$\displaystyle x^3\sin y - y^3\cos{\color{red}x} \:=\:x^3$
We have: .$\displaystyle x^3\cos y\!\cdot\!\frac{dy}{dx} + 3x^2\sin y - y^3(\text{-}\sin x) - 3y^2\cos x\!\cdot\!\frac{dy}{dx} \:=\:3x^2 $
. . . . . . $\displaystyle x^3\cos y\!\cdot\!\frac{dy}{dx} - 3y^2\cos x\!\cdot\!\frac{dy}{dx} \:=\:3x^2-3x^2\sin y - y^3\sin x$
Factor: .$\displaystyle \left(x^3\cos y - 3y^2\cos x\right)\,\frac{dy}{dx} \:=\:3x^2-3x^2\sin y - y^3\sin x$
Therefore: .$\displaystyle \frac{dy}{dx} \;=\;\frac{3x^2-3x^2\sin y - y^3\sin x}{x^3\cos y - 3y^2\cos x} $