## Find dy/dx by using implicit differentiation: sinx=x(1+tany)

Limits, differentiation, related rates, integration, trig integrals, etc.

### Find dy/dx by using implicit differentiation: sinx=x(1+tany)

Find dy/dx by using implicit differentiation: sinx=x(1+tany)

Here's what I have.

dy/dx[sinx]=dy/dx[x(1+tany)]
cosx=x[(sec^2y)dy/dx]+(1+tany)
cosx-tany-1=dy/dx[x(sec^2y)]
dy/dx=(cosx-tany-1)/(xsec^2y)

I haven't done this in a while so I am not confident in my answer. Did I do it right?
Cafe au lait

Posts: 6
Joined: Mon Jan 03, 2011 11:02 pm

### Re: Find dy/dx by using implicit differentiation: sinx=x(1+t

Looks good to me. Thank you for showing your work so nicely!
nona.m.nona

Posts: 255
Joined: Sun Dec 14, 2008 11:07 pm

### Re: Find dy/dx by using implicit differentiation: sinx=x(1+t

Cafe au lait wrote:dy/dx[sinx]=dy/dx[x(1+tany)]

your notation here is not correct

$\frac{d}{dx}[\sin(x)]=\frac{d}{dx}[x(1+\tan(y))]$

Martingale

Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: Find dy/dx by using implicit differentiation: sinx=x(1+t

Ooh, thanks for catching that Martingale! Fixed that on my paper.
Cafe au lait

Posts: 6
Joined: Mon Jan 03, 2011 11:02 pm