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
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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