1 + x = sin(xy^2)
find dy/dx by implicit differentiation
0 + 1 = cos(xy^2). (x)(2y)dy/dx + (y^2)(1)
1/((x)(2y)dy/dx) = cos(xy^2) + (y^2)
dy/dx = cos (xy^2) + (y^2)....
Can I just divide out the (x)(2y) and leave the dy/dx?
you needed brackets like this
0 + 1 = cos(xy^2)*( (x)(2y)dy/dx + (y^2)(1) )
since you need to get at the dy/dx and since it is inside the bracket, you must expand
1 = 2xy(cos(xy^2))dy/dx + y^2(cos(xy^2)
1 - y^2(cos(xy^2) = 2xy(cos(xy^2))dy/dx
dy/dx = (1 - y^2(cos(xy^2))/(2xy(cos(xy^2)))
No, you cannot simply divide out the (x)(2y) term and leave the dy/dx term. When you have an equation like 1 + x = sin(xy^2), and you want to find dy/dx using implicit differentiation, you need to follow these steps:
1. Start by differentiating both sides of the equation with respect to x.
- Differentiating 1 + x with respect to x gives us 0 + 1 = 0 + dx/dx = 1.
- Differentiating sin(xy^2) with respect to x requires the use of the chain rule. Remember that sin(u), where u is a function of x, has the derivative du/dx * cos(u). So, we have dy/dx * (x * (2y))^' * cos(xy^2). Applying the chain rule, we get dy/dx * (2y) * x' * cos(xy^2), which simplifies to dy/dx * 2xy * cos(xy^2).
2. Combine the terms involving dy/dx on one side of the equation and other terms on the other side.
- In this case, we have 1 = dy/dx * 2xy * cos(xy^2).
3. Solve for dy/dx.
- To isolate dy/dx, divide both sides of the equation by 2xy * cos(xy^2).
- The final expression for dy/dx is dy/dx = 1 / (2xy * cos(xy^2)).
Note that you cannot simply divide out the (x)(2y) term because it is already involved in the derivative of sin(xy^2) with respect to x. Implicit differentiation requires treating dy/dx as a separate variable and applying the chain rule appropriately.