1. ## [SOLVED] Differentiation

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?

2. Originally Posted by xwrathbringerx
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

3. 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: .$\displaystyle \frac{dy}{dx} \;=\;\frac{3x^2-3x^2\sin y - y^3\sin x}{x^3\cos y - 3y^2\cos x}$