Implicit differentiation help?
In general the procedure is as follows: Let F(x,y) = 0. The total differential of F is: ∂F/∂x dx + ∂F/∂y dy = 0 Therefore you find the derivative of y with respect to x by: dy/dx = - ∂F/∂x / ∂F/∂y Here: F(x,y) = tanh(x/y) - artanh(y/x) you get the partial derivatives by applying the chain rule: ∂F/∂x = d( tanh(x/y)) / d(x/y) · ∂(x/y)/∂x + d( artanh(y/x)) / d(y/x) · ∂(y/x)/∂x = sech²(x/y) · 1/y - 1/(1- (y/x)²) · y/x² (that's what you have calculated ) ∂F/∂y = d( tanh(x/y)) / d(x/y) · ∂(x/y)/∂y + d( artanh(y/x)) / d(y/x) · ∂(y/x)/∂y = sech²(x/y) · x/y² - 1/(1- (y/x)²)· 1/x Thus: dy/dx = [sech²(x/y) · 1/y - 1/(1 - (y/x)²)· y/x²] / [sech²(x/y) · x/y² - 1/(1- (y/x)²)· 1/x] = [sech²(x/y) · (x/y) - (x/y)²/(1- (y/x)²) ] / [sech²(x/y) · (x/y)² - 1/(1- (y/x)²)]
Implicit Differentiation Help?
Implicit derivatives look like this. The derivative of 5y = 5 dy. The derivative of -2 is 0. (-2 is a constant.) To get the derivative of 3xy use the product rule. First factor times derivative of second factor plus second factor times derivative of first factor. The first factor is 3x and the second factor is y. So the derivative is 3x dy + y(3 dx) = 3x dy + 3y dx. For the sin(2x + 3y), let u = 2x+3y. du = 2 dx + 3 dy. So we are finding the derivative of sin u. This will be cos u du = cos(2x+3y)(2 dx + 3 dy). Put the derivatives together: cos(2x+3y)(2 dx + 3dy) = 3x dy + 3y dx + 5 dy. Distribute the left side: 2 cos(2x+3y) dx + 3 cos(2x+3y) dy = 3x dy + 3y dx + 5 dy. Divide all terms by dx. 2cos(2x+3y)dx/dx. Dx cancels. 3cos(2x+3y)dy/dx. 3x dy/dx. 3y dx/dx (dx cancels here also.) 5 dy/dx. 2cos(2x+3y) + 3cos(2x+3y) dy/dx = 3x dy/dx + 3y + 5 dy/dx. 3cos(2x+3y) dy/dx -3x dy/dx -5 dy/dx = 3y - 2cos(2x+3y) Factor dy/dx and divide both sides. dy/dx = (3y - 2 cos(2x+3y))/(3cos(2x+3y) -3x -5) Phew!
Implicit differentiation???? HELP?
d/dx(cosxsiny)=0 -sinxsiny+cosxcosy(dy/dx)=0 dy/dx=(sinxsiny)/(cosxcosy) =tanxtany
Implicit Differentiation help ...?
find dr/dt of sin(rt) = 1/3 I just can't figure it out. I thought it was r(cos(rt)) but apparently that isn't right .... and find dy/dx of e^(2x) = sin(x+6y) I'm trying to finish up my homework and study for a quiz tomorrow and I have been stuck on these two questions for two days. any help would be soooo appreciated.
Just need help to get started- Partial differential Equations?
I've been working on this problem for a few hours now, and can't seem to finish it up (get started) The thing that is throwing me off is the au(b-u). What exactly does this represent. How can I get started? Derive an equation for the concentration u(x,t) of a chemical pollutant if the chemical is produced due to chemical reaction at the rate of au(b-u) per unit volume. Thanks so much!
Need Help With Implicit Differentiation?
For implicit differentiation you will need at least the to use the chain rule and product rule: d(y^n)/dx = (n-1)y^(n-1)*dy/dx or (n-1)y^(n-1)*y' 4x2 + 9y2 = 36 8x + 18y*y' = 0 That is it, now solve explicitly for y' y' = - 8x/(18y) = - 4 x/(9y) If you put y' in terms of x you will have y' = ± something First solve for y: 4x2 + 9y2 = 36 9y^2 = 36 - 4x^2 y^2 =( 36 - 4x^2)/9 y = ±√(36 - 4x^2)/3 substitute and we get: y' = ±4x/[3√(36 - 4x^2)]