# Implicit Differentiation

• Apr 21st 2009, 02:28 AM
Russ
Implicit Differentiation
I was working though my problem book for Calculus and I was able to do the simple implicit differentiation questions such as:

$x^2$y = 1

(I don't know how to do derivatives in Latex, somebody help!)

d/dx ( $x^2$y) = d/dx (1)

( $x^2$)d/dx(y) + y(2x) = 0

( $x^2$)(dy/dx) + 2xy = 0

Rearrange and you get:

dy/dx = -2y/x

So I could do that and I get the basic idea of what we're doing, but there were two questions that really stumped me.

(a) xy = tan (xy)

(b) $x^2$ = (arctan(y)) / (1 + $y^2$)

I am completely lost, any help would be muchly appreciated :)

EDIT: for (a) I did manage to get a line that said:

(x)(dy/dx) + y = sec(xy)

That on the right track?
• Apr 21st 2009, 03:07 AM
Mush
Quote:

Originally Posted by Russ
I was working though my problem book for Calculus and I was able to do the simple implicit differentiation questions such as:

$x^2$y = 1

(I don't know how to do derivatives in Latex, somebody help!)

d/dx ( $x^2$y) = d/dx (1)

( $x^2$)d/dx(y) + y(2x) = 0

( $x^2$)(dy/dx) + 2xy = 0

Rearrange and you get:

dy/dx = -2y/x

So I could do that and I get the basic idea of what we're doing, but there were two questions that really stumped me.

(a) xy = tan (xy)

(b) $x^2$ = (arctan(y)) / (1 + $y^2$)

I am completely lost, any help would be muchly appreciated :)

EDIT: for (a) I did manage to get a line that said:

(x)(dy/dx) + y = sec(xy)

That on the right track?

First of all, once you get to $\frac{dy}{dx} = \frac{-2y}{x}$, you should write $y$ in terms of $x$ by rearranging the original equation. So $y = \frac{1}{x^2}$, therefore

$\frac{dy}{dx} = \frac{-2\bigg(\frac{1}{x^2}\bigg)}{x} = \frac{-2x^{-2}}{x} = \frac{-2}{x^3}$

Righty.

Fir the tan one, remember that you must use the chain rule if the argument is not exactly x. Your LHS is fine. But:

$\frac{d}{dx} tan(xy) = \sec^2(xy) \times \frac{d}{dx} (xy) = y\sec^2(xy)$
• Apr 21st 2009, 03:15 AM
Russ
Oh really? Okay. I can do that from now on.

Hmmm, i'm not sure why I didn't have a $sec^2 (xy)$ in there
Thanks I will work on this some more
• Apr 21st 2009, 03:22 AM
Mush
Quote:

Originally Posted by Russ
Oh really? Okay. I can do that from now on.

Hmmm, i'm not sure why I didn't have a $sec^2 (xy)$ in there
Thanks I will work on this some more

For the last one:

$x^2 = \frac{\arctan(y)}{1+y^2}$

I would start by multiply both sides by $1+y^2$, which gives@

$x^2(1+y^2) = \arctan(y)$

Hence:

$x^2 + x^2y^2 = \arctan(y)$

For the RHS:

$\frac{d}{dx} \arctan(y) = \frac{dy}{dx} \frac{d}{dy} \arctan(y)$

And $\frac{d}{dy} \arctan(y) = \frac{1}{1+y^2}$
• Apr 21st 2009, 03:25 AM
Russ
Ah nice, thanks. I think I was on track for that, but I was trying to use the quotient rule and things were getting complicated.
• Apr 23rd 2009, 02:42 AM
mauriziocorso77
Quote:

Originally Posted by Mush

$\frac{d}{dx} tan(xy) = \sec^2(xy) \times \frac{d}{dx} (xy) = y\sec^2(xy)$

Incorrect. the differentiation of xy is y+xy'. Therefore $(tan(xy))' = (sec^2(xy))(y+xy')$