implicit differentiation con't ...01

**Foxlion** · March 10th 2011, 07:26 PM

Problem:

$4cosxsiny = 1$

Solution attempt:

create differentiation schema;
$siny[\frac{d}{dx}4(cosx)+4\frac{d}{dx}(cosx)] + 4cosx[\frac{d}{dx}siny]=\frac{d}{dx}1$

apply differentiation schema;
$siny[-4sinx]+4cosx[cosyy'] = 0$

simplify;
$-4sinxsiny+4cosxcosyy' = 0$

add $4sinxsiny$ to both sides;
$4cosxcosyy' = 4sinxsiny$

divide both sides by $4cosxcosyy'$ ;
$y' = \frac{4sinxsiny}{4cosxcosy}$

apply trig identity $\frac{sinx}{cosx} = tanx$ and simplify;
$y' = tanxtany$

Answer has been verified, thank you all.

**topsquark** · March 10th 2011, 07:36 PM

Originally Posted by Foxlion

Problem:

$4cosxsiny = 1$

Solution attempt:

create differentiation schema;
$siny[\frac{d}{dx}4(cosx)+4\frac{d}{dx}(cosx)] + 4cosx[\frac{d}{dx}siny]=\frac{d}{dx}1$

apply differentiation schema;
$siny[-sinx]+4cosx[cosxy'] = 0$

There are two problems with this line. You dropped a factor of 4 (that's just a typo, I think.) The other is the inclusion of an x instead of a y in the second term. It should read
$4siny[-sinx]+4cosx[cosyy'] = 0$

-Dan

**Foxlion** · March 10th 2011, 07:50 PM

Originally Posted by topsquark

It should read
$4siny[-sinx]+4cosx[cosyy'] = 0$

-Dan

Done and done though I am still lost on steps 7 and 8. Is it necessary to expand with the trig idents?

**NOX Andrew** · March 10th 2011, 08:17 PM

Originally Posted by Foxlion

apply differentiation schema;
$siny[-4sinx]+4cosx[cosyy'] = 0$

simplify and factor;
$-4sinxsiny+y'(4cos^2y) = 0$

I believe there is an error in this step, namely re-writing $4 \cos x \cos y$ as $4 \cos ^2 y$ .

**Foxlion** · March 10th 2011, 08:24 PM

ah yes as dan's correction eliminates the existential possibility of $cos^2x$ I'm such a fool.

original post edited for corrections again thank you all.

**topsquark** · March 11th 2011, 03:07 PM

Originally Posted by Foxlion

Problem:

$4cosxsiny = 1$

Solution attempt:

create differentiation schema;
$siny[\frac{d}{dx}4(cosx)+4\frac{d}{dx}(cosx)] + 4cosx[\frac{d}{dx}siny]=\frac{d}{dx}1$

apply differentiation schema;
$siny[-4sinx]+4cosx[cosyy'] = 0$

simplify;
$-4sinxsiny+4cosxcosyy' = 0$

add $4sinxsiny$ to both sides;
$4cosxcosyy' = 4sinxsiny$

divide both sides by $4cosxcosyy'$ ;
$y' = \frac{4sinxsiny}{4cosxcosy}$

apply trig identity $\frac{sinx}{cosx} = tanx$ and simplify;
$y' = tanxtany$

Answer has been verified, thank you all.

Please do not alter the original question. (See Forum Rule #7 here.) We all do some amount of it, but the amount of editing you have done here makes it hard to see what you did and why all these posts exist. It can get very confusing.

-Dan

**Foxlion** · March 11th 2011, 08:01 PM

Originally Posted by topsquark

Please do not alter the original question. We all do some amount of it, but the amount of editing you have done here makes it hard to see what you did and why all these posts exist. It can get very confusing.

-Dan

It can, I apologize. From now on I shall leave the original intact save for purely typographical corrections.

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