Hi,
I'm having trouble using implicit differentiation to determine dy/dx in the form dy/dx = f(x,y) for
sin(2x+3y)=3x^3y^2+4
Do I make it sin(2x+3y)-3x^3y^2=4 then differentiate to get
2cos(2x+3y)-9x^2*2y=0 ?
I'm a little lost...
Any help appreciated.
No. Your first step is IK but unecessary, and your second differentiation step is wrong.
Differentiate both sides of the equation with repect to x, remembering that y is a function of x. Use the "chain rule" when necessary.
cos(2x+3y)*(2 + 3dy/dx) = 6y^2x^2 +6x^3*y*dy/dx
Rearrange and solve for dy/dx
dy/dx[3cos(2x+3y)-6x^3 y]
= -2cos(2x+3y)- 6y^2x^2
One more step
Thanks for your reply.
I tried again and got
cos(2x+3y)*(2 + 3dy/dx = 9x^2y^2 + (6x^3*y dy/dx)
Looking at your answer it seems wrong also.
Should it not be 9y^2x^2... on the 1st part of the right hand side of your equation? Or am I missing something here?
Getting stuck on the rearranging but will it another try
Yes, it's 9x^2 y^2 where I had a coefficient of 6. Good work!
Awesome. Think I've done it!
Cheers for the help.
Hi,
Implicit differentiation can be tricky at times, but let me explain step-by-step how to approach this problem.
1. Start by differentiating both sides of the equation with respect to x. Keep in mind that y is considered as a function of x.
So, we have:
d/dx(sin(2x + 3y)) = d/dx(3x^3y^2 + 4)
2. Apply the chain rule on the left side of the equation. Since we have sin(2x + 3y), we need to differentiate the outer function (sin) and the inner function (2x + 3y).
Differentiating the outer function:
d/dx(sin(u)) = cos(u) * du/dx
Differentiating the inner function:
d/dx(2x + 3y) = 2 + 3 * dy/dx
So, rewriting the left side of the equation using the chain rule:
cos(2x + 3y) * (2 + 3 * dy/dx) = d/dx(3x^3y^2 + 4)
3. Now, reorganize the equation to solve for dy/dx. Move all the terms involving dy/dx to one side of the equation, and move the other terms to the other side.
cos(2x + 3y) * 3 * dy/dx = d/dx(3x^3y^2 + 4) - 2 * cos(2x + 3y)
4. Simplify the terms on the right side of the equation. Since 4 is a constant, its derivative with respect to x is 0. And since the derivative of x^3y^2 with respect to x involves both x and y, we need to apply the product rule.
d/dx(3x^3y^2 + 4) = 9x^2y^2 * dy/dx + 6x^2y * dx/dx = 9x^2y^2 * dy/dx + 6x^2y
So, substituting the simplified terms on the right side:
cos(2x + 3y) * 3 * dy/dx = 9x^2y^2 * dy/dx + 6x^2y - 2 * cos(2x + 3y)
5. Now, isolate dy/dx by moving all the terms involving dy/dx to one side of the equation.
cos(2x + 3y) * 3 * dy/dx - 9x^2y^2 * dy/dx = 6x^2y - 2 * cos(2x + 3y)
6. Factor out dy/dx.
dy/dx * (cos(2x + 3y) * 3 - 9x^2y^2) = 6x^2y - 2 * cos(2x + 3y)
7. Finally, solve for dy/dx by dividing both sides of the equation by (cos(2x + 3y) * 3 - 9x^2y^2).
dy/dx = (6x^2y - 2 * cos(2x + 3y)) / (cos(2x + 3y) * 3 - 9x^2y^2)
And there you have it! You have found the derivative dy/dx in terms of x and y.
I hope this explanation helps you understand the steps involved in using implicit differentiation to find dy/dx for the given equation. Let me know if you have any further questions!