find y' if x^2y^4+3y-4x^3=5cos x-1
(2x*y^4)+dy/dx 4y^3*x^2)+dy/dx3+12x^2=5-sin-0
dy/dx=-5sin (x -12 x^2)/(2x*4y^3)+(12y^2+x^2)
this is using implicit differenation is this correct
d/dx of
x^2y^4+3y-4x^3=5cos x-1
((( note - dx/dx = 1)))
so
x^2[4y^3dy/dx] +y^4[2x]-12x^2 = -5sin x
4x^2y^3 dy/dx = 12x^2 -2xy^4-5sinx
dy/dx=3/y^3 - y/2x- 5sinx/(4x^2y^3)
(2x*y^4)+y'4y^3*x^2)+ 3y'+12x^2=-5sinx
On that line I am not certain if the original problems was 5cos(x-1) or 5cos x -1. I assume the latter. Then gathering terms..
y'(4x^2y^3+3)=(-5sinx -2xy^4-12x^2)
I do not see this leading to a y' which is the same as your manipulations.
x^2[4y^3dy/dx] +y^4[2x]+3 -12x^2 = -5sin x
4x^2y^3 dy/dx = 12x^2 -2xy^4-5sinx -3
dy/dx=3/y^3-y/2x-(3+5sinx)/(4x^2y^3)
x^2[4y^3dy/dx] +y^4[2x]+3dy/dx -12x^2 = -5sin x
[4x^2y^3+3]dy/dx = 12x^2 -2xy^4-5sinx
dy/dx
=(12x^2 -2xy^4-5sinx)/[4x^2y^3+3]
Yes, your calculation is correct. You have applied the implicit differentiation correctly to find the derivative dy/dx. This technique allows us to find the derivative of an equation that is not explicitly given as y = f(x). In this case, the given equation is x^2y^4 + 3y - 4x^3 = 5cos(x) - 1.
To find the derivative, you have implicitly differentiated each term with respect to x and then found dy/dx.
The derivative of x^2y^4 with respect to x is obtained by using the product rule, which states that (uv)' = u'v + uv'. Here, u = x^2 and v = y^4. Applying the product rule, we get (x^2y^4)' = (2x^2y^4) + (x^2)'(y^4).
Similarly, applying the product rule to the other terms, we have (3y)' = (3)(y)' = 3(dy/dx) and (-4x^3)' = (-4)(x^3)' = -12x^2.
The derivative of 5cos(x) - 1 with respect to x is given by the derivative of each term individually, which is -5sin(x).
After differentiating each term and rearranging the equation, you correctly solved for dy/dx as:
dy/dx = (-5sin(x) - 12x^2) / (2x * 4y^3) + (12y^2 + x^2)
So, your result is correct. Good job!