How Do You Solve These Complex Differentiation Problems?

[billmccai]

Homework Statement

Find dy/dx for:

a) xsin(xy^2) - ln(x/y) = y

b) x = 3cos(a), y = 2sec(a). simplify then find dy/dx when a = pi/3

Homework Equations

The Attempt at a Solution

I've done both of them, was just hoping someone could check I've done things correctly. I'm still feeling pretty uncertain.


a)

xsin(xy^2) - ln(x/y) = y

dy/dx = d/dx[xsin(xy^2) - ln(x/y)]

dy/dx = sin(xy^2) + (xcos(xy^2) * (y^2 + 2xy*dy/dx)) - 1/(x/y) * (1/y - x/(y^2) *dy/dx)

dy/dx = sin(xy^2) + xy^2cos(xy^2) + 2x^2ycos(xy^2)dy/dx - (1/y)/(x/y) + (x/y^2)/(x/y) * dy/dx

thus

dy/dx = (-sin(xy^2) - xy^2cos(xy^2) +1/x) / (2x^ycos(xy^2) + 1/y - 1)



b)

dy/dx = (dy/da) / (dx/da)

= 2sec(a)tan(a) / -3sin(a)

using tan(a) = sin(a) / cos(a), and sec(a) = 1/cos(a)

dy/dx = -(2/3) * 1/cos(a) * sin(a)/cos(a) * 1/sin(a)

= -(2/3) * (1/cos^2(a))

so for a = pi/3

dy/dx = -(2/3) * (1/ (.5^2))
= -(2/3) * 4
= -2 2/3

 

[arildno]

There might be some sign mistakes lurking in there that I didn't catch, but your general procedure is sound.

At b), why not just use:
$$\cos(a)=\frac{x}{3}\to\sec(a)=\frac{3}{x}\to{y}=\frac{6}{x}\to\frac{dy}{dx}=-\frac{6}{x^{2}}$$
Then, a=pi/3 implies x=3/2, whereby $$\frac{dy}{dx}=-\frac{8}{3}$$