# Thread: Calculus Problems, finding dy/dx.

1. ## Calculus Problems, finding dy/dx.

Can someone help me with any of these problems? I'll really appreciate it!

12) Complete the curve defined by x^4+4x^2y+y^4=1 find dy/dx

13) For the curve (x+1)/(y+1)=2 dy/dx= [(y+1)(2y+1)]/(x+1)
a) if dy/dx= [(y+1)(2y+1)]/(x+1) Find the equation of the tangent line to the curve at (3,1)

b) if dy/dx= [(y+1)(2y+1)]/(x+1) write the equation(s)of any horizontal or vertical tangent line(s) OR show work to explain why there are none

14) if dy/dx= x^2y^2 find d^2y/dx^2 as an expression of x and y only.You do not need to simplify.

2. ## Re: Calculus Problems, finding dy/dx.

Note first that by the chain rule d/dx(y^4)=d/dy(y^4) times dy/dx=4y^3dy/dx
Also note a term like x^2y is a product and will require the product rule to differentiate it
In your first question differentiate each term (BOTH sides) with respect to x
Get 4x^3+(8xy+4x^2dy/dx)+4y^3dy/dx=0 The part in brackets came from using the product rule)
So (4x^2+4y^3)dy/dx=-4x^3-8xy dy/dx=-4(x^3+2xy)/4(x^2+y^3)=-(x^3+2xy)/(x^2+y^3)
The process wew have used here is called implicit differentiation.