# Thread: implicit differentiation con't ...01

1. ## implicit differentiation con't ...01

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.

2. 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

3. 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?

4. 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 $4 \cos ^2 y$.

5. 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.

6. 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

7. 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.