I previously posted this but had a mistake in it.
Determine dy/dx if xsiny=arctany
i did (x)(cosy dy/dx)+(1)(siny)=dy/(1+y^2) so far
To find dy/dx, we need to differentiate both sides of the equation. Let's break down the steps:
Step 1: Differentiate xsiny with respect to x
To find the derivative of xsiny, we use the product rule. The product rule states that if we have a function u(x)v(x), then the derivative is given by u'(x)v(x) + u(x)v'(x).
For xsiny, let's identify u(x) and v(x):
u(x) = x
v(x) = siny
Differentiating u(x) and v(x):
u'(x) = 1 (since the derivative of x is 1)
v'(x) = cosy (since the derivative of siny with respect to x is cosy)
Applying the product rule:
(x)(cosy) + (1)(siny)
Step 2: Simplify the right side of the equation
As you have correctly done, the derivative of arctany with respect to x is given by dy/(1+y^2).
Step 3: Equate the two expressions and solve for dy/dx
Set the left side of the equation equal to the right side and solve for dy/dx:
(x)(cosy) + siny = dy/(1+y^2)
Now, you can solve the equation for dy/dx by isolating the dy/dx term on one side of the equation.