Need help with understanding implicit differentiation problem?
No, y is equal to 100 for the specific value of x = 4, not for all x. First forget about the fact that y(4) is 100. Differentiate the equation with respect to x d (sqrt(x) + sqrt(y) ) / dx = d(12)/dx d (sqrt(x))/dx + d (sqrt(y))/dx = d(12)/dx d (sqrt(x))/dx + d (sqrt(y))/dx = 0 (derivative of a constant is 0) 1/(2 sqrt(x) ) + d (sqrt(y))/dx = 0 1/(2 sqrt(x) ) + d (sqrt(y))/dy * dy/dx = 0 (by using the chain rule of the second term) 1/(2 sqrt(x) ) + 1/(2 sqrt (y) * dy/dx = 0 1/(2 sqrt(x) ) + 1/(2 sqrt (y) * y' = 0 Now we have the condition: y(4) = 100, which means that for x = 4, y = 100 So we plug thowse values in the equation: 1/(2sqrt(4)) + 1/(2 sqrt(100)) * y'(4) = 0 1/4 + y'(4) /20 = 0 y'(4) = -20/4 = -4 Hope this helps
Math help! please implicit differentiation?
√x + √y = √c => 1/(2√x) + 1/(2√y) dy/dx = 0 => dy/dx = - √y/√x Let (x', y') be any point on the curve => equation of the tangent at that point is y - y' = - (√y'/√x') (x - x') x-intercept of this tangent is obtained by plugging y = 0 => 0 - y' = - (√y'/√x') (x - x') => x = √(x'y') + x' y-intercept of the tangent is obtained by plugging x = 0 => y - y' = - (√y'/√x') (0 - x') => y = y' + √(x'y') Sum of the x and y intercepts = √(x'y') + x' + y' + √(x'y') = (√x' + √y')^2 = (√c)^2 (because (x', y') is on the curve => √x' + √y' = √c) = c.
Help with implicit differentiation?
The graph of the equation x^2+xy+y^2 = 9 is a slanted ellipse illustrated in the figure http://webwork2.asu.edu/webwork2_course_files/Reynolds_MAT_265_Fall_2010/tmp/gif/Section_2.6-prob11-fig.gif Think of y as a function of x. Differentiating implicitly and solving for y' gives: y' = ___--- i got this to be (-2x-y)/(x+2y), and that is correct but i dont know how to solve the rest. The ellipse has two horizontal tangents. The upper one has the equation: y = ___ The right most vertical tangent has the equation: x = ___ That tangent touches the ellipse where: y = ___ If you can help please do!
Find dy/dx by implicit differentiation. sqrt (7xy) = 9 + x^2y?
sqrt (7xy) = 9 + x^2y (7xy)^(1/2) = 9+x^2 y differentiate both sides with respect to x (1/2)(7xy)^(-1/2) d/dx[xy] = 0 + 2xy + x^2 y' (1/2)(7xy)^(-1/2) sqrt(7) (y + x y') = 2xy + x^2 y' sqrt(7) /2sqrt(7xy) (y+ xy') = 2xy + x^2 y' 1/2sqrt(xy) (y + xy') = 2xy + x^2 y' y'( x/2sqrt(xy) -x^2) = 2xy - y/2sqrt(xy) y' = ( 2xy - y/2sqrt(xy)) / (x/2sqrt(xy) -x^2) --------------------------------------... Note: y/2sqrt(xy) = y over 2sqrt(xy) x/2sqrt(xy) = x/2sqrt(xy)
How would you differentiate the equation [math]x^2 + 4xy + y^2=20[/math]?
Assuming that you are looking for the derivative [math]\dfrac{dy}{dx}[/math]. Taking the derivative of both sides with respect to x, we have[math]\dfrac{d}{dx}[x^2+4xy+y^2]=\dfrac{d}{dx}[20][/math][math]\implies 2x + 4(y+xy’)+2yy’=0[/math][math]\implies x + 2y + 2xy’+yy’=0[/math] [Dividing throughout by [math]2[/math]][math]\implies y’=-\dfrac{x+2y}{2x+y}[/math][math]\implies \dfrac{dy}{dx}=-\dfrac{x+2y}{2x+y}[/math]
How would I differentiate dy/dx with respect to t?
Perhaps an example would help.Assume that [math]x^2+y^2 = 1[/math] (unit circle)Assume that this circle is a track or road and that a vehicle (ball?) is moving along it, and that x and y represent where the vehicle/ball/whatever is at a particular time.Then both x and y are functions of t (time)By implicit differentiation we can find that[math]\displaystyle 2x + 2 y \frac{dy}{dx}=0[/math]and solve for [math]\frac{dy}{dx}[/math][math]\displaystyle\frac{dy}{dx}=-\frac{x}{y}[/math]The question wants[math]\displaystyle\frac{d}{dt}\frac{dy}{dx}[/math]This means, how fast the slope is changing at a particular timeTo answer the question, we can remove y from the formulas or use the chain rule in its two variable form:[math]\displaystyle\frac{d}{dt}(-\frac{x}{y}) = \frac{\partial}{\partial{x}}(-\frac{x}{y})\frac{dx}{dt} + \frac{\partial}{\partial{y}}(-\frac{x}{y})\frac{dy}{dt}[/math]If we have both x and y as explicit functions, then the situation simplifies considerably.For example, suppose the angular velocity is constant at [math]\omega[/math], meaning that[math]\displaystyle{x}=\cos(\omega t)[/math][math]\displaystyle{y}=\sin(\omega t)[/math]Then our first result above gives us that[math]\displaystyle\frac{dy}{dx}=-\frac{x}{y} = -\tan(\omega t)[/math]So now we can find the time derivative very easily.
How do I differentiate √1-x^2?
Change the eqn to another form [math](1-x^2)^{1/2}[/math]Use chain rule.[math]\dfrac{dy}{dx}=\dfrac{1}{2}(1-x^2)^{-1/2}(-2x)[/math][math]\dfrac{dy}{dx}=-\dfrac{2x}{2\sqrt{1-x^2}}[/math][math]\dfrac{dy}{dx}=-\dfrac{x}{\sqrt{1-x^2}}[/math]
Calculus: implicit differentiation/particle's distance changing question?
A particle is moving along the parabola 4y = (x + 2)^2 in such a way that its x-coordinate is increasing at a constant rate of 2 units per second. How fast is the particle's distance to the point (-2, 0) changing at the moment that the particle is at the point (2, 4)?
Question on differentiation please help!!!!?
a) y = (kx - x^2)^(1/2) y' = (1/2)(kx - x^2)^(-1/2)(k - 2x) when x = 2, y' = 2^(1/2) (1/2)(k*2 - 2^2)^(-1/2)(k - 2*2) = 2^(1/2) (1/2)(2k - 4)^(-1/2)(k - 4) = 2^(1/2) k - 4 = 2(2k - 4)^(1/2)(2^(1/2)) square both sides k^2 - 8k + 16 = 16k - 32 k^2 - 24k + 48 = 0 k = 12 + 4sqrt(6) = 21.8 b) the slope of the normal is -1/sqrt(2) when x = 2 y = (21.8*2 - 2^2)^(1/2) y = 6.3 the equation of the normal is (y - 6.3)/(x - 2) = -1/sqrt(2) -sqrt(2)y + 6.3sqrt(2) = x - 2 x + sqrt(2)y = -6.3sqrt(2) - 2 c = -6.3sqrt(2) - 2 = -10.9
If sqrt(x) + sqrt(y) = 12 ...and y(25)=49, find y'(25) by implicit differentiation.?
apply derive d/dx , d/dx(sqrt(x) +d/dx(sqrt(y)= d/dx 12 (1/2)x^(-1/2) +(1/2)y(-1/2) dy/dx =0 dy/dx = -( x/y)^(-1/2) or dy/dx= (y/x)^(1/2) at x=25 , y(25)= 49 dy/dx = (49/25)^(1/2) = +- 7/5 Exists onother way to do this , using total differentiation If F(x,y) =Constant , then dF= DF/dx dx +dF/dy dy =0 D ( Partial derive ) , then, dF/dx = DF/dx +DF/dy dy/dx =0 dy/dx = -(DF/dx)/(DF/dy) In your case, DF/dx = (1/2) x^(-1/2) , DF/dy = (1/2)y^(-1/2) dy/dx = -( x/y)^(-1/2) The same.- by this method you can also obtain dx/dy ( Do DF/dy) or dx/dt ( in parametric equations) ( Do DF/dt)