Question

find dy/dx by implicit differentiation. tan(x-y)=y/(7+x^(2))

Answer 1

Oh boy this one is going to be a little messy. I dunno about you, but I really dislike using the quotient rule, so let's multiply that denominator to the other side.\[\large \left(7+x^2\right)\tan(x-y)=y\]

From here, we'll take the derivative implicitly, applying the product rule on the left.

Answer 2

\[\large \color{royalblue}{\left(7+x^2\right)'}\tan(x-y)+\left(7+x^2\right)\color{royalblue}{\tan(x-y)'}=\color{royalblue}{y'}\]
So here is the setup for the product rule, we'll have to take the derivative of eveyrthing in blue. Understand how I set that up? :O

Answer 3

yes.

Answer 4

I got 2x*tan(x-y)+(7+x^2)*sec^2(x-y)=y'

Answer 5

Ok good, you just missed one simple step. You forget to apply the `Chain Rule` to the inside of the tangent.

Answer 6

oh, so would it be 2x*tan(x-y)*y'+(7+x^2)*sec^2(x-y)=y'?

Answer 7

Woops wrong tangent, the chain rule will be applied to the tangent that you took a derivative of, look at this to help maybe.
\[\large (2x)\tan(x-y)+\left(7+x^2\right)\sec^2(x-y)\color{royalblue}{(x-y)'}=y'\]

Answer 8

do you multiply by y' in the product rule when you take the derivative of something with a y term?

Answer 9

Yes, kind of.
The problem is, the inside of our tangent function wasn't simply `y`.
You're probably used to dealing with problems like \(\large \tan y\) , yes in that case a y' would pop out when you take it's derivative.

But this is due to the chain rule.
You'll only have a y' when you take the derivative of y directly.
I dunno, maybe that's a bad way to explain it.

\[\large \frac{d}{dx}\tan y \qquad = \qquad \sec^2y(y)'\]The chain rule tells us to multiply by the derivative of the inside, so we'll do that,\[\large = \qquad (\sec^2y)y'\]
----------------------------

Our problem is a little trickier.\[\large \frac{d}{dx}\tan(x-y) \qquad =\qquad \sec^2(x-y)(x-y)'\]The derivative of the inside gives us,\[\large =\qquad \sec^2(x-y)(1-y')\]

Answer 10

so then the product rule would be 2x*tan(x-y)+(7+x^2)*sec^2(x-y)*(1-y')=y'?

Answer 11

After applying the product rule, we apply the chain rule to the second term, and yes that's what we get.

Now comes a bit of tricky algebra. Moving stuff around, shuffle shuffle, try to solve get the y' alone on one side.

Answer 12

that's where I get lost with implicit differentiation.

Answer 13

I know you get the y' terms on one side together

Answer 14

Do you understand how to get the y' on one side? Or need to see some steps? c:

Answer 15

I need to see some steps.. not good with algebra..

Answer 16

\[\large (2x)\tan(x-y)+\color{orangered}{\left(7+x^2\right)\sec^2(x-y)}(1-y')=y'\]
Let's start by distributing the big... orange mess of stuff... into that brackets containing y'
\[\large (2x)\tan(x-y)+\color{orangered}{\left(7+x^2\right)\sec^2(x-y)}-\color{orangered}{\left(7+x^2\right)\sec^2(x-y)}y'=y'\]

Answer 17

We can then move the y' term over to the right side by adding it to both sides,\[\large (2x)\tan(x-y)+\left(7+x^2\right)\sec^2(x-y)=y'+\left(7+x^2\right)\sec^2(x-y)y'\]

Answer 18

Then factor a y' ouch of each term.\[\large (2x)\tan(x-y)+\left(7+x^2\right)\sec^2(x-y)=y'\left[1+\left(7+x^2\right)\sec^2(x-y)\right]\]

Answer 19

Then divide by the big mess...\[\large \frac{(2x)\tan(x-y)+\left(7+x^2\right)\sec^2(x-y)}{\left[1+\left(7+x^2\right)\sec^2(x-y)\right]}=y'\]

Answer 20

We end up with something really awful, like that.... :\

Answer 21

Thank you for showing what you did AND explaining the step because that's helpful for me. And thank you for taking the time to help! Also, the one thing I am unsure of is the distributing step.

Answer 22

We took our giant ugly potato and distributed it (multiplied it by each term) inside the brackets. As illustrated here.
What makes it confusing maybe is that it's a really big potato. So it's hard to understand that we're distributing that entire thing to each term. It almost looked like we made a copy of it, which maybe looked weird :)