Derivative implicit differentiation problem

[laylas]

Homework Statement

Consider the curve satisfying the equation 2(x+1)^(tanx)=(y^2)cosx+y and find dy/dx

Homework Equations

(tanx)'=sec^2x
(lnx)'=1/x

The Attempt at a Solution

I've tried taking the natural log of both sides and then taking d/dx of both sides but something seems to go wrong with my algebra each time.. that or I'm getting confused about my natural log / implicit differentiation rules. I'm not actually sure if I'm approaching the problem correctly but here's one of the many attempts I've made:

ln2 + tanxln(x+1) = 2lny + lncosx + lny(??)
1/2 + sec^2xln(x+1) + tanx(1/x+1) = 2y'/y + sin(x)/cos(x) + y'/y
something should already be wrong so i won't post the rest of the work but i know the answer is supposed to be dy/dx = [2(x+1)^(tanx)[sec^2xln(x+1)+(tanx)/(x+1)]+(y^2)(sinx)]/ all over (2ycosx+1). Need help getting there, thanks!

 

[Strants]

It looks like you tried to distribute the natural log over addition, which doesn't work. The left hand side is fine, but the right hand side should be
\ln (y^2 \cos(x) + y)
Or I suppose you could divide out the y from the original expression, but that won't really help, seeing as you'd just have to combine the two fractions you get after differentiating.

 

[Dick]

The hard part for you seems to be differentiating (x+1)^tan(x). f(x)^g(x)=e^(ln(f(x))*g(x)) since f(x)=e^(ln(f(x)). Try differentiating it in that form.

 

[laylas]

Once I have the right side down to 1/y^2(cosx+y) [?] how do I find dy/dx?

 

[SammyS]

Once I have the right side down to 1/y^2(cosx+y) [?] how do I find dy/dx?

If you have the right side down to " 1/y^2(cosx+y) " ... Where did dy/dx go, and what is in the numerator & what in the denominator?

By the way: 1/y^2(cosx+y) literally is the same as (cosx+y)/y² . It helps to use enough parentheses to say what you mean