Trying to find dy/dx of a trig function

[jtt]

Homework Statement

find dy/dx

Homework Equations

x= tan y

The Attempt at a Solution

d/dx(x=tanY)

1=(tan(dy/dx))+sec^2+y

 

[Mark44]

Homework Statement

find dy/dx

Homework Equations

x= tan y

The Attempt at a Solution

d/dx(x=tanY)

First off, you don't take the derivative of an equation - you take the derivative of each side of an equation.

The above should be d/dx(x) = d/dx(tan(y))

To carry out the differentiation on the right, you need to use the chain rule.

d/dx(f(u)) = d/du(f(u)) * du/dx = f'(u) * du/dx

1=(tan(dy/dx))+sec^2+y

Nope.