Dull Differentiation

Calculus Level 1

Let y y be a function of x x such that x 3 + y 3 = 4 x^{3} + y^{3} = 4 .

Find y ′ = d y d x y' = \dfrac {dy}{dx} .

− x 2 y 2 \frac{-x^2}{y^2} x 2 y 2 \frac{x^2}{y^2} y 2 − x 2 \frac{y^2}{-x^2} − y 2 − x 2 \frac{-y^2}{-x^2}

Rajdeep Ghosh by Rajdeep Ghosh , 18, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Chew-Seong Cheong
Aug 28, 2016

x 3 + y 3 = 4 Differentiate both sides w.r.t x 3 x 2 + 3 y 2 d y d x = 0 3 y 2 d y d x = − 3 x 2 d y d x = − 3 x 2 3 y 2 ⟹ y ′ = − x 2 y 2 \begin{aligned} x^3 + y^3 & = 4 & \small \color{#3D99F6}{\text{Differentiate both sides w.r.t }x} \\ 3x^2 + 3y^2 \frac {dy}{dx} & = 0 \\ 3y^2 \frac {dy}{dx} & = - 3x^2 \\ \frac {dy}{dx} & = \frac {-3x^2}{3y^2} \\ \implies y' & = \boxed{\dfrac {-x^2}{y^2}} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Rajdeep Ghosh
Aug 28, 2016

Begin with x 3 + y 3 = 4 x^{3} + y^{3} = 4 . Differentiate both sides of the equation, getting

3 x 2 + 3 y 2 y ′ = 0 , 3x^2 + 3y^2 y' = 0 ,

so that (Now solve for y' .)

3 y 2 y ′ = − 3 x 2 , 3y^2 y' = - 3x^2 ,

and

y ′ = − 3 x 2 3 y 2 = − x 2 y 2 . y' = \displaystyle{ - 3x^2 \over 3y^2 } = \displaystyle{ - x^2 \over y^2 } .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

x 3 + y 3 = 4 x^3+y^3=4

Differentiate with respect to x x .

3 x 2 + 3 y 2 d y d x = 0 3x^2 + 3y^2\dfrac{dy}{dx}=0

3 y 2 d y d x = − 3 x 2 3y^2\dfrac{dy}{dx}=-3x^2

d y d x = − 3 x 2 3 y 2 = − x 2 y 2 \dfrac{dy}{dx}=\dfrac{-3x^2}{3y^2}=\boxed{\dfrac{-x^2}{y^2}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...