# Implicit Differentiation Problems

• Oct 20th 2009, 03:28 PM
helpplz
Implicit Differentiation Problems
4 problems I can't figure out:

1) (x^2)(y^2) - y = x
2) 2sinx(cosy) = 1
3) sinx = x(1+tany)
4) y = sin(xy)

Please don't simplify or try to isolate y from the beginning. =)
Only simplify to isolate dy/dx
• Oct 20th 2009, 03:33 PM
skeeter
Quote:

Originally Posted by helpplz
4 problems I can't figure out:

1) (x^2)(y^2) - y = x

product rule with the $\displaystyle \textcolor{red}{x^2y^2}$ term

2) 2sinx(cosy) = 1

product rule again ...

3) sinx = x(1+tany)

product rule on the right ...

4) y = sin(xy)

product rule with the xy argument

what seems to be your problem with using the product rule for derivatives?
• Oct 20th 2009, 03:38 PM
helpplz
I'm not sure why I just don't know where to start..I think I just don't know how to apply product rule with the y's..
• Oct 20th 2009, 03:41 PM
skeeter
Quote:

Originally Posted by helpplz
I'm not sure why I just don't know where to start..I think I just don't know how to apply product rule with the y's..

(1st function)(derivative of the 2nd) + (2nd function)(derivative of the 1st)

like so ...

$\displaystyle \frac{d}{dx}(x^2y^2) = x^2 \cdot 2y \frac{dy}{dx} + y^2 \cdot 2x$
• Oct 20th 2009, 03:43 PM
helpplz
Hm, but why is it 2y(dy/dx) instead of just 2(dy/dx)?
• Oct 20th 2009, 03:46 PM
skeeter
Quote:

Originally Posted by helpplz
Hm, but why is it 2y(dy/dx) instead of just 2(dy/dx)?

because it was $\displaystyle y^2$ to begin with ... not $\displaystyle 2y$. you still have to use the power rule.
• Oct 20th 2009, 03:49 PM
helpplz
Quote:

Originally Posted by skeeter
because it was $\displaystyle y^2$ to begin with ... not $\displaystyle 2y$. you still have to use the power rule.

So if it was something like y^3, the derivative would be 3y(dy/dx)^2 ?
• Oct 20th 2009, 03:56 PM
skeeter
Quote:

Originally Posted by helpplz
So if it was something like y^3, the derivative would be 3y(dy/dx)^2 ?

no

$\displaystyle \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$
• Oct 20th 2009, 04:02 PM
helpplz
Quote:

Originally Posted by skeeter
no

$\displaystyle \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$

Ah, okay, gotcha. I'm gonna do a random problem just to make sure I'm doing it right:

x^2y^2 -2x = 3

x^2(2y(dy/dx)) + y^2(2x) = 5

dy/dx = 5-y^2(2x)/ x^2(2y)

That right? (Wondering)
• Oct 20th 2009, 04:28 PM
skeeter
Quote:

Originally Posted by helpplz
Ah, okay, gotcha. I'm gonna do a random problem just to make sure I'm doing it right:

x^2y^2 -2x = 3

x^2(2y(dy/dx)) + y^2(2x) = 5

not quite ...

$\displaystyle \textcolor{red}{x^2 \cdot 2y \frac{dy}{dx} + y^2 \cdot 2x - 2 = 0}$

the derivative of a constant is 0.
• Oct 20th 2009, 04:41 PM
helpplz
Quote:

Originally Posted by skeeter
the derivative of a constant is 0.

Oh geez, I must have forgotten that on at least 15 problems...Thanks sooo much for correcting me. =) And for all your help!