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

[Cafe au lait]

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?

 

[galactus]

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

Yep, you got it. Good show

You could also write it as:

$\displaystyle y'=\frac{(cos(x)cos(y)-cos(y)-sin(y))cos(y)}{x}$

But I suppose it really does not matter.

$\displaystyle y'=\frac{cos(x)-tan(y)-1}{xsec^{2}(y)}$ is equivalent.

Wouldn't think so, huh?. :wink:

 

[Cafe au lait]

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

Thank you for your help! Your answers are formatted very nicely.