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Find dy/dx in terms of \( x \& y \) : \( \begin{array}{l}\text { 1. } x y-x+2 y=1 \\ \text { 2. } x^{3}+y^{3}=1 \\ \text { 3. } x^{2}+x y=y^{3} \\ \text { 4. } x^{3} y+x y^{5}=2 \\ \text { 5. } x^{2} y^{3}=2 x-y \\ \text { 6. } x^{2}+4(y-1)^{2}=4 \\ \text { 7. } \frac{x-y}{x+y}=\frac{x^{2}}{y}+1\end{array} \)

Ask by Reyes Barnett. in Canada
Dec 16,2024

Real Tutor Solution

Tutor-Verified Answer

Answer

\( \sqrt{2} \).

Solution

Calculate or simplify the expression \( \sin(45) \). Calculate the value by following steps: - step0: Calculate: \(\sin\left(45\right)\) Calculate or simplify the expression \( \cos(\pi/4) \). Calculate the value by following steps: - step0: Calculate: \(\cos\left(\frac{\pi }{4}\right)\) - step1: Calculate the trigonometric value: \(\frac{\sqrt{2}}{2}\) Calculate or simplify the expression \( \sin(45)/\cos(\pi/4) \). Calculate the value by following steps: - step0: Calculate: \(\frac{\sin\left(45\right)}{\cos\left(\frac{\pi }{4}\right)}\) - step1: Transform the expression: \(\frac{2\sin\left(45\right)}{\sqrt{2}}\) - step2: Multiply by the Conjugate: \(\frac{2\sin\left(45\right)\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}\) - step3: Multiply the numbers: \(\frac{2\sqrt{2}\times \sin\left(45\right)}{\sqrt{2}\times \sqrt{2}}\) - step4: Multiply the numbers: \(\frac{2\sqrt{2}\times \sin\left(45\right)}{2}\) - step5: Reduce the fraction: \(\sqrt{2}\times \sin\left(45\right)\) The value of \( \frac{\sin(45^{\circ})}{\cos(\frac{\pi}{4})} \) is \( \sqrt{2} \).

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Sure! Let’s explore how to differentiate these implicit functions step by step. For the first equation, \( xy - x + 2y = 1 \), you can implicitly differentiate both sides with respect to \( x \). Applying the product rule and moving terms involving \( dy/dx \) will give you the result: \[ y + x \frac{dy}{dx} - 1 + 2 \frac{dy}{dx} = 0 \] Rearranging gives you \( \frac{dy}{dx} = \frac{1 - y}{x + 2} \). Now let’s jump to the fifth equation, \(x^2y^3 = 2x - y\). Again, use implicit differentiation: \[ 2xy^3 + 3x^2y^2\frac{dy}{dx} = 2 - \frac{dy}{dx} \] After rearranging and solving for \( \frac{dy}{dx} \), you would find \[ \frac{dy}{dx} = \frac{2 - 2xy^3}{3x^2y^2 + 1} \] And there we have it! You're ready to tackle these implicit derivatives like a pro.

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