Solution for 9 If y cf(x), where c is a constant and (x) is a differentiable function prove that dy/dx = cf'(x) %3DCalculus Find dy/dx y=1/x y = 1 x y = 1 x Differentiate both sides of the equation d dx (y) = d dx ( 1 x) d d x ( y) = d d x ( 1 x) The derivative of y y with respect to x x is y' y ′ y' y ′ Differentiate the right side of the equation Tap for more stepsM(x,y)dxN(x,y)dy= 0 is defined implicitly by φ(x,y)= c, where φ satisfies (194) and c is an arbitrary constant Proof We rewrite the differential equation in the form M(x,y)N(x,y) dy dx = 0 Since the differential equation is exact, there exists a potential function φ (see (194)) such that ∂φ ∂x ∂φ ∂y dy dx = 0 But this
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Find dy/dx if x^y y^x=c where c is constant- Best answer Given, xy = yx Taking logarithm on both sides, we get y log x = x log y Differentiating both sides, wrt x y (1/x) log x (dy/dx) = x (1/y) (dy/dx) log y 1 (y/x) (log x) (dy/dx) = (x/y) (dy/dx) log y (dy/dx) log x (x/y) = log y (y/x) (dy/dx) (y log x x)/ySolution Given differential equation is y d y d x x = c ⇒ y d y = ( c − x) d x On integrating both sides, we get y 2 2 = c x − x 2 2 d ⇒ y 2 x 2 − 2 c x − 2 d = 0 Hence, it represents a family of circles whose centres are on the x axis
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World of mathematics Something went wrong Wait a moment and try againClick here 👆 to get an answer to your question ️ If y = c , a constant then find dy/dx rutvij4873 rutvij4873 Math Secondary School If y = c , a constant then find dy/dx 2 It follows that 1/t = dy/dxlnx> x^y(lnx(dy/dx) y/x) Doing the same, we get that y^x(lny x/y(dy/dx)) Thus the derivative of the entire function is given by x^ylnx(dy/dx) x^y(y/x) y^xln y y^x x/y(dy/dx) = 0 x^ylnx(dy/dx) y^x x/y(dy/dx) = x^y(y/x) y^xlny dy/dx = (x^y(y/x) y^xln y)/(x^ylnx y^x (x/y)) Hopefully this helps!
2 Answers2 Active Oldest Votes 3 d ( y c) = d y d c However, if c is a constant, then d c = 0 so we get d ( y c) = d y Consequently, if one side of the following makes sense, then both sides do and they are equal A general solution of the differential equation \((x y) \frac{dy}{dx} = x y\) will be _____ where c is a constant Q15 The solution of the differential equation \((yx)^2 \dfrac{dy}{dx}=a^2\) is given bySolution For Find (dy)/(dx)\ if (x^2y^2)^2=x y
Dy/dx = x y, which we can think of as slope = x y But isoclines are where slope is constant So let's rewrite the equation again as constant = x y Or, more briefly as x y = C, where C represents any constant A little algebra, (very little), gives us y = x C, A very common integration problem is when we are given an expression for `dy/dx` and we are given a point that the curve passes through and we need the find the original function This is integration because we are told the `dy/dx` and we need to find the original expression Ok, so if I've got `dy/dx` equals 3x 2 2 I recognise that what I need to do is to integrate thisThe derivative of tan(x) tan ( x) with respect to x x is sec2(x) sec 2 ( x) sec2(x) sec 2 ( x) Reform the equation by setting the left side equal to the right side y' = sec2(x) y ′ = sec 2 ( x) Replace y' y ′ with dy dx d y d x dy dx = sec2(x) d y d x = sec 2 ( x)
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If xy=c 2, then y=c 2 /x So y/x = c 2 /x 2 You got the same answer, just in a different formKCET 15 If x = ct and y = (c/t) , find (dy/dx) at t=2 (A) (1/4) (B) 4 (1/4) (D) 0 Check Answer and Solution for above question from MathNotice there is no 0th order derivative here Hence, this is actually just a firstorder equation in disguise Substitute v = \frac{dy}{dx} \frac{dv}{dx} = v^2 Separate this and solve v(x) = \frac{1}{c
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The rate at which y is changing is dy/dx times the rate at which x is changing (that's a rigorous statement) Which is pretty much the same thing as saying that dy/dx is dy divided by dx But you have to keep in mind that this is only a heuristic!Dy/dx = u(dv/dx) v(du/dx) Now, from y = 3xsinx, u will be = 3x, and v = sinx du/dx = 3, and dv/dx = cosx, hence, dy/dx = 3x(cosx) sinx(3) dy/dx = 3sinx 3xcosx Now for the right answer to the above question Option A is incorrect Option B is incorrect C is incorrect D is the correct answer KEYPOINTS You may please note these/this Find a continuous solution satisfying $$ \frac{dy}{dx} y= f(x) $$ Where $$ f(x) = \begin{cases} 1 &\text{ for } 0 < x < 1, \\ 0 & \text{ for } x > 1 \end{cases} $$ w Stack Exchange Network Stack Exchange network consists of 177 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn
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Transcribed image text If xy = C, where is a constant, then the corresponding differential equation is α" ₂4 dy dx = 4xy 423 4 y=0 dy 2 dx 4x4y = 0 x ₂4 dy hodiny 4x®y = 0(Write your answer in the form F(x,y) = C, where C stands for an arbitrary constant) {eq}\frac{dy}{dx} = \sqrt4{(\frac{x}{y})} ,or, \frac{dy}{dx} = (\frac{x}{y})^{\frac{1}{4}} {/eq}Answer d //dx=y1/1x Given the equation xy=xy We need to find dy/dx We will use implicit differentiation ==> xy'=xy'
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Example 35 Find 𝑑𝑦/𝑑𝑥 , if 𝑥 = 𝑎𝑡2, 𝑦 = 2𝑎𝑡 𝑑𝑦/𝑑𝑥 = 𝑑𝑦/𝑑𝑥 × 𝑑𝑡/𝑑𝑡 𝑑𝑦/𝑑𝑥 = 𝑑𝑦/𝑑𝑡 × 𝑑𝑡/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Calculating 𝒅𝒚/𝒅𝒕 𝑦 = 2𝑎𝑡" "𝑑𝑦/𝑑𝑡 " " = 𝑑(2𝑎𝑡)/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 2𝑎 𝑑(𝑡)/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝟐𝒂Calculating 𝒅𝒙/𝒅𝒕 Prachi Agarwal answered this x = ct dx/dt = d/dt ( ct ) = c * d/dt ( t) t * d/dt (c) applying product rule = c (1) t (0) difft of t is 1 and difft of constant c is 0 = c ie, dx/dt = c now, y = cWe have the following theorem Theorem 1
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Uploaded By Army_Dillo Pages 6 This preview shows page 2 4 out of 6Now suppose we find the derivative of y with respect to a, but TREAT x as the constant Then dy da = x Here we just "reversed" the roles played by a and x in our equation Example 11 For f(x,y,z) use the implicit function theorem to find dy/dx and dy/dz 1 f(x,y,z)=x 2y3 z xyz Example 24 Find 𝑑𝑦/𝑑𝑥 if 𝑥 − 𝑦 = 𝜋 (x − y) = 𝜋 Differentiating both sides wrt x (𝑑(𝑥 − 𝑦))/𝑑𝑥= 𝑑𝜋/𝑑𝑥 𝑑𝑥/𝑑𝑥− 𝑑𝑦/𝑑𝑥 = 0 1 − 𝑑𝑦/𝑑𝑥 = 0 𝒅𝒚/𝒅𝒙 = 1 (As 𝜋 is constant)
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(Use C as an arbitrary constant) Variable Separable Equation Any differential equation that can be written in the form, {eq}h\left( x \right)dx = k\left( y \right)dy {/eq}, is a differentialCourse Title COST ACCOU 1;RHS x log(2) => log(2) log(2) is a constant so x dissapears So we get (1/y)(dy/dx) = log(2) 4) We want to find dy/dx, which is on the LHS To get this dy/dx on its own we can multiply both sides by y So we get dy/dx = y log(2) 5) To finish this question we need to sub in for y and then we have an answer for dy/dx
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X c Cancelling the logarithm from both sides of the above equation, we get y = x c This is the required solution of the given differential equation ⇐ Solve Differential Equation dy/dx=xe^y ⇒ Separable Variables of Differential Equations ⇒Solution of ` (x y)^2 (dy)/ (dx)=a^2` ('a', being a constant) is solution of ` (x y)^2 (dy)/ (dx)=a^2` ('a', being a constant) is Watch later Share Copy link Info Shopping And now i am clueless how to find C Becouse I cannot find arctan pi/4 and there is a problem with pi^2/161 I assume that C=0 becouse graphic of arctan x and 1/(x^21) are very close in point 079 (pi/4), but i cannot prove it
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If x y y x = a b, find dy/dx cbse;Find dy dx y c cos x x 2 sin x c is constant 2 pts A c cos x x cos x sin x C c Find dy dx y c cos x x 2 sin x c is constant 2 pts a School Ateneo de Davao University;(x 3y)dx (x 2y)dy = 0 Existence of a solution The general solution of the equation dy/dx = g(x, y), if it exists, has the form f(x, y, C) = 0, where C is an arbitrary constant Under what circumstances does a general solution exist?
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If x=at2 and y=at, then find dy/dx?If x^2y^2 =25 and dy,dt =6, find dx,dt when y=4Solution for dx/dy px = q, where p and q are functions of y or constants A differential equation is called linear if there are no multiplications among dependent variables and their derivatives In other words, all coefficients are functions of independent variables It can be written in two forms The second is discussed in detail here
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Answer x^(y1) y y^x logy/1 x^y logx x y^(x1) Solution y = x^y y^x Let u = x^y and v = y^x Then, y = u v and dy/dx = du/dx dv/dx(1) We will now find du/dx and dv/dx separately and substitute iThis might appear to have no solutions given y(0) = 1, but as it turns out the constant y= 1 is itself a solution 5 dy dx = e yx;y(0) = 2 e y dy= ex dx e y = c ex e y = c ex y= ln( c ex) y= ln(1 1=e ex) Note that this solution will not be de ned for all values of x, since ln(x) is only de ned for positiveIn Introduction to Derivatives (please read it first!) we looked at how to do a derivative using differences and limits Here we look at doing the same thing but using the "dy/dx" notation (also called Leibniz's notation) instead of limits We start by calling the function "y" y = f(x) 1 Add Δx When x increases by Δx, then y increases by Δy
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We can neglect C ′ C' C ′ since another integral constant C C C is hidden in the integral ∫ f (x) d x \int f(x)~dx ∫ f (x) d x Therefore, we can conclude that if d y d x = f (x), \frac{dy}{dx}=f(x), d x d y = f (x), then y = ∫ f (x) d x C y=\int f(x)~dxC y = ∫ f (x) d x C Find y y y in terms of x x x where d y d x = xClick here👆to get an answer to your question ️ If y^x = x^y , then find dy/dxFind an answer to your question if x^y y^x = c find dy by dxplease give answer with simple steps
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Find y 0 so that the integral curve for dy/dx=xy y(4)=y 0 is a straight line You must justify your answer, which will require you to apply algebraic reasoning to the problem I know the answer is 3 I also know this differential equation looks simple, but IShare It On Facebook Twitter Email 1 Answer 1 vote answered by paayal (147k points) selected by Vikash Kumar Best answer Put x y = u and y x = v ∴ (i) becomesHere we will look at solving a special class of Differential Equations called First Order Linear Differential Equations First Order They are "First Order" when there is only dy dx, not d 2 y dx 2 or d 3 y dx 3 etc Linear A first order differential equation is linear when it can be made to look like this dy dx P(x)y = Q(x) Where P(x) and Q(x) are functions of x To solve it there is a
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