1. implicit differentiation

Given:
$\displaystyle ye^y=(x-5y)^3$

Find $\displaystyle \frac{dy}{dx}$, in terms of x and y.

what I tried to do was

Differentiate implicitly wrt x,
$\displaystyle \frac{dy}{dx}(e^y)+\frac{dy}{dx}(y)(e^y)=(3)(x-5y)^2(1-5\frac{dy}{dx})$
$\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2(1-5\frac{dy}{dx})$

then, i'm stuck in moving all $\displaystyle \frac{dy}{dx}$ to the left-hand side.
any help is much appreciated. thanks.

2. Re: implicit differentiation

Originally Posted by wintersoltice
Given:
$\displaystyle ye^y=(x-5y)^3$

Find $\displaystyle \frac{dy}{dx}$, in terms of x and y.

what I tried to do was

Differentiate implicitly wrt x,
$\displaystyle \frac{dy}{dx}(e^y)+\frac{dy}{dx}(y)(e^y)=(3)(x-5y)^2(1-5\frac{dy}{dx})$
$\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2(1-5\frac{dy}{dx})$

then, i'm stuck in moving all $\displaystyle \frac{dy}{dx}$ to the left-hand side.
any help is much appreciated. thanks.
$\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2-15\frac{dy}{dx}(x-5y)^2$

$\displaystyle \frac{dy}{dx}[e^y(1+y)+15(x-5y)^2]=3(x-5y)^2$

$\displaystyle \frac{dy}{dx}=\frac{3(x-5y)^2}{e^y(1+y)+15(x-5y)^2}$

3. Re: implicit differentiation

Originally Posted by alexmahone
$\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2-15\frac{dy}{dx}(x-5y)^2$

$\displaystyle \frac{dy}{dx}[e^y(1+y)+15(x-5y)^2]=3(x-5y)^2$

$\displaystyle \frac{dy}{dx}=\frac{3(x-5y)^2}{e^y(1+y)+15(x-5y)^2}$
how did $\displaystyle (1-5\frac{dy}{dx})$
become $\displaystyle -15\frac{dy}{dx}(x-5y)^2$?

4. Re: implicit differentiation

Originally Posted by wintersoltice
how did $\displaystyle (1-5\frac{dy}{dx})$
become $\displaystyle -15\frac{dy}{dx}(x-5y)^2$?
$\displaystyle 3(1-5\frac{dy}{dx})=3-15\frac{dy}{dx}$