1. implicit differentiation

find $\displaystyle dy/dx$ by implicit differentiation

$\displaystyle \tan^3{(xy^2+y)} = 1$

didn't know how to deal with the $\displaystyle \tan^3$

$\displaystyle \frac {1-3y^2\tan^2(xy^2+y)\sec^2(xy^2+y)} {3(2xy+1)tan^2(xy^2+y)sec^2(xy^2+y)}$

2. Hello,

For the differentiation of tan^3, use the chain rule !

Here are the first steps :

We know that the derivative of $\displaystyle \tan^3(x)=3\cdot\tan^2(x)\cdot \frac{1}{\cos^2(x)}$, by the chain rule.

So the derivative of $\displaystyle \tan^3(x)(z)$, where $\displaystyle z=xy^2+y$ will be $\displaystyle 3\cdot\tan^2(z)\cdot \frac{1}{\cos^2(z)}\cdot\frac{dz}{dx}$, by the chain rule.

Now I bet you can differentiate $\displaystyle z=xy^2+y$ with respect to x (not forgetting that y is a function of x)

3. Originally Posted by bigwave
find $\displaystyle dy/dx$ by implicit differentiation

$\displaystyle \tan^3{(xy^2+y)} = 1$

didn't know how to deal with the $\displaystyle \tan^3$

$\displaystyle \frac {1-3y^2\tan^2(xy^2+y)\sec^2(xy^2+y)} {3(2xy+1)tan^2(xy^2+y)sec^2(xy^2+y)}$
Hi,

Just recall the chain rule... first you take the power and forget everything else:

$\displaystyle thing^3(stuff) = 1$

Now we differentiate:

$\displaystyle 3thing^2(stuff)thing' = 1$

Now, we must differentiate the stuff inside, so that we get:

$\displaystyle 3*thing^2(stuff)*(thing)'*(stuff)' = 1$

So, if we let:

$\displaystyle thing = tan()$
$\displaystyle stuff = xy^2 + y$
$\displaystyle thing' = sec^2()$
$\displaystyle stuff' = y^2 + 2xyy' + y'$

Plug this all in to get:

$\displaystyle 3tan^2(xy^2 + y)[sec^2(xy^2 + y)(y^2 + 2xyy' + y')] = 1$

Now, we can divide everything without a y' to the other side:

$\displaystyle y^2 + 2xyy' + y' = \frac{1}{3tan^2(xy^2 + y)sec^2(xy^2 + y)}$

Now, we move the y squared over and factor out the y':

$\displaystyle y'(2xy + 1) = \frac{1}{3tan^2(xy^2 + y)sec^2(xy^2 + y)} - y^2$

Now we divide through by 2xy + 1:

$\displaystyle y' = \frac{1}{3(2xy + 1)tan^2(xy^2 + y)sec^2(xy^2 + y)} - \frac{y^2}{2xy + 1}$

And if you get common denominators it becomes the answer you have... Hope this helps.

4. also, can we consider $\displaystyle tan^3(x)$ the same as $\displaystyle (tan(x))^3$ to apply $\displaystyle d/dx$

5. Originally Posted by bigwave
how did you get this?

I thot $\displaystyle \tan^3{(x)} = 3\tan^2{(x)}\sec^2{(x)}$

also, can we consider $\displaystyle tan^3(x)$ the same as $\displaystyle (tan(x))^3$ to apply $\displaystyle d/dx$
Sorry, I edited while you were apparently posting. Got messed up with some derivatives

As for your second sentence, yes, $\displaystyle \tan^3(x)=(\tan(x))^3$